. 

*.  J. 


DABBY    V 


INSTITUTE 

DAUBY  WISKE. 


University  of  California  •  Berkeley 

THE  THEODORE  P.  HILL  COLLECTION 

of 
EARLY  AMERICAN  MATHEMATICS  BOOKS 


I 


TREATISE 


ON 


ALGEBRA. 


BY  ELIAS  LOOMIS,  A.M., 

FBOFESSOB  OP  MATHEMATICS  AND  NATURAL  PHILOSOPHY  IN  THE  UNIVERSITY  OF  THE 

CITY  OF  NEW   YORK,   MEMBER   OF  THE  AMERICAN  PHILOSOPHICAL   SOCIETY, 

OF   THE  AMERICAN  ACADEMY  OF  ARTS  AND   SCIENCES,  AND 

AUTHOR  OF  "ELEMENTS  OF  GEOMETRY." 


THIRD      EDITION. 


NEW    YORK: 

HARPER   &  BROTHERS,   PUBLISHERS, 

82   CLIFF   STREET. 

1850, 


Entered,  according  to  Act  of  Congress,  in  the  year  one  thousand 
eight  hundred  and  forty-seven,  by 

HARPER  &  BROTHERS, 

in  the  Clerk's  Office  of  the  District  Court  of  the  Southern  District 
of  New  York. 


TO 


THE    COUNCIL 


OF 


THE  UNIVERSITY  OP  THE  CITY  OP  NEW  YORK, 


THIS    TREATISE 


respectfully 


THEIR  OBEDIENT  SERVANT, 


THE  AUTHOR. 


PR  E  F  ACE. 


THE  first  edition  of  my  Algebra  was  received  with  unex- 
pected favor.  Almost  immediately  after  its  publication,  it  was 
adopted  as  a  text-book  in  half  a  dozen  colleges,  besides  nu- 
merous academies  and  schools ;  and  the  most  flattering  testi- 
monials were  received  from  every  part  of  the  country.  I 
have  thus  been  stimulated  to  increased  exertions  to  render  it 
less  unworthy  of  public  favor.  Every  line  of  it  has  been  sub- 
jected to  a  thorough  revision.  The  work  has  been  read  by 
two  successive  classes  in  the  University,  and  wherever  im- 
provement seemed  practicable,  alterations  have  been  freely 
made.  I  have  also  availed  myself  of  the  suggestions  of  sev- 
eral professors  in  other  colleges.  This  edition  will  accord- 
ingly be  found  to  differ  considerably  from  the  preceding.  Al- 
terations, more  or  less  important,  have  been  made  on  nearly 
every  page.  Among  these  may  be  mentioned  the  addition  of 
Continued  Fractions,  the  Extraction  of  the  Roots  of  Numbers, 
Elimination  by  means  of  the  Greatest  Common  Divisor,  and  a 
large  collection  of  Miscellaneous  Examples. 

It  is  believed  that  this  treatise  contains  as  much  of  Algebra 
as  can  be  profitably  read  in  the  time  allotted  to  this  study  in 
most  of  our  colleges,  and  that  those  subjects  have  been  se- 
lected which  are  most  important  in  a  course  of  mathematical 
study.  These  materials  I  have  endeavored  to  combine,  so  as 
to  form  a  consistent  treatise.  I  have  aimed  to  cultivate  in  the 
mind  of  the  student  a  habit  of  generalization,  and  to  lead  him  to 


IV  PREFACE. 

reduce  every  principle  to  its  most  general  form.  At  the  same 
time,  I  have  been  solicitous  not  to  discourage  the  young  begin- 
ner, who  frequently  finds  it  much  more  difficult  to  comprehend 
a  general  than  a  particular  proposition.  Accordingly,  many 
of  the  Problems  have  been  twice  stated.  I  first  give  a  simple 
numerical  problem,  and  then  repeat  the  same  problem  in  a 
more  general  form.  I  have  labored  to  develop,  in  a  clear 
and  intelligible  manner,  the  most  important  properties  of  equa- 
tions, and  have  bestowed  great  pains  upon  the  selection  of 
examples  to  illustrate  these  properties.  Throughout  the  work 
I  have  endeavored  to  render  the  most  important  principles  so 
prominent  as  to  arrest  attention ;  and  I  have  reduced  them,  as 
far  as  practicable,  to  the  form  of  concise  and  simple  rules.  It 
is  believed  that,  in  respect  of  difficulty,  this  treatise  need  not 
discourage  any  youth  of  fifteen  years  of  age  who  possesses 
average  abilities,  while  it  is  designed  to  form  close  habits  of 
reasoning,  and  cultivate  a  truly  philosophical  spirit  in  more 
mature  minds. 


CONTENTS. 


SECTION  I. 

DEFINITIONS  AND  NOTATION. 

DEFINITIONS. — Difference  between  Algebra  and  Arithmetic 1 

Signs  of  Addition  and  Subtraction 4 

Signs  of  Multiplication  and  Division 5 

Coefficient. — Exponent. — Power. — Boot 6 

Algebraic  Quantity.— Monomial. — Polynomial 8 

Degree  of  a  Term. — Homogeneous  Polynomials 9 

SECTION  II. 

ADDITION  .  .  12 


SECTION  III. 

SUBTRACTION 16 

SECTION  IV. 

MULTIPLICATION. 

Case  of  Monomials. — Rule  for  the  Exponents 21 

Case  of  Polynomials. — Rule  for  the  Signs 23 

Degree  of  a  Product. — Number  of  Terms  in  a  Product 26 

Theorems  proved. — Quantities  resolved  into  Factors 27 

Multiplication  by  detached  Coefficients 29 

SECTION  V. 

DIVISION. 

Case  of  Monomials. — Rule  of  Exponents 32 

Negative  Exponents. — Symbol  aP 33 

Case  of  Polynomials 36 

a"  _  Jn  Divisible  by  a  —  b 39 

Quantities  resolved  into  Factors 40 

Division  by  detached  Coefficients 41 

SECTION  VI. 

FRACTIONS. 

Fundamental  Principles. — Signs  of  the  Terms 43 

To  reduce  a  Fraction  to  lower  Terms 45 

To  reduce  a  Fraction  to  an  entire  or  mixed  Quantity 4G 


vj  CONTENTS. 

Page 

To  reduce  a  mixed  auantity  to  the  Form  of  a  Fraction '• 

To  reduce  Fractions  to  a  common  Denominator < 

Addition  and  Subtraction  of  Fractions « 

Multiplication  and  Division  of  Fractions ! 

SECTION  VII. 

SIMPLE   EQUATIONS. 

Definitions.— Axioms  employed ! 

Equations  solved  by  Subtraction  and  Addition ( 

Equations  solved  by  Division  and  Multiplication < 

Equations  cleared  of  Fractions ( 

Solution  of  Problems • '" 65 


SECTION  VIII. 

EQUATIONS  WITH   TWO   OR  MORE  UNKNOWN  QUANTITIES. 

Elimination  by  Substitution 78 

By  Comparison.— By  Addition  and  Subtraction 79 

Equations  containing  three  unknown  Quantities - 86 

Equations  containing  m  unknown  Quantities 88 

SECTION  IX. 

DISCUSSION   OF  EQUATIONS   OF  THE   FIRST   DEGREE. 

Positive  Values  of  a;.— Negative  Values - 92 

Infinite  Values.— Indeterminate  Values 96 

Zero  and  Infinity • 

Inequalities 10° 

SECTION  X. 

INVOLUTION    AND   POWERS. 

Involution  of  Monomials.— Sign  of  the  Result 105 

Involution  of  Polynomials 108 

t 

SECTION  XI. 

EVOLUTION  AND  RADIPAL  QUANTITIES. 

To  extract  a  Hoot  of  a  Monomial 110 

Sign  of  the  Root. — Square  Root  of  a  Trinomial 112 

Irrational  Quantities. — Fractional  Exponents 115 

To  reduce  Surds  to  their  most  simple  Forma 117 

To  reduce  a  Rational  Quantity  to  the  Form  of  a  Surd 120 

To  reduce  Surds  to  a  common  Index 121 

Addition  and  Subtraction  of  Surd  Quantities 122 

Multiplication  and  Division  of  Surd  Quantities 124 

Involution  and  Evolution  of  Surd  Quantities 127 

To  find  Multipliers  which  shall  render  Surds  Rational 129 

To  reduce  a  Surd  Fraction  to  a  Rational  Numerator  or  Denominator 131 

To  free  an  Equation  from  Radical  Quantities 133 

Calculus  of  Imaginary  Expressions ,,.,  135 


CONTENTS.  Vll 

SECTION  XII. 

EQUATIONS  OF  THE   SECOND  DEGREE. 

Page 

Solution  of  Pure  Quadratics 137 

Solution  of  Complete  Quadratics 142 

Solution  of  the  Equation  aft"  -\-pxn  =  g 148 

Quadratic  Equations  containing  two  unknown  Quantities 156 

Discussion  of  the  general  Equation  of  the  Second  Degree 161 

Discussion  of  particular  Problems 168 

Double  Values  of  a— Imaginary  Values 169 

SECTION  XIII. 

RATIO  AND   PROPORTION. 

Definitions. — Ratios  compared  with  each  other 171 

Proportion. — Product  of  Extremes  and  Means I75 

Equal  Ratios. — Alternation. — Inversion - 178 

Composition. — Division. — Conversion I79 

Like  Powers  of  Proportionals.—  Continued  Proportion 180 

Harmonical  Proportion. — Variation 182 

SECTION  XIV. 

PROGRESSIONS. 

Arithmetical  Progression. — L  ast  Term. — Sum  of  the  Terms 188 

Geometrical  Progression. — L  ast  Term. — Sum  of  the  Terms 194 

Progressions  having  an  infinite  Number  of  Terms 198 

Harmonical  Progres sion 201 

SECTION  XV. 

GREATEST   COMMON  DIVISOR. — CONTINUED   FRACTIONS. — PERMUTATIONS  AND   COM- 
BINATIONS. 

Greatest  common  Divisor. — How  found 203 

Rule  applied  to  Polynomials 205 

Continued  Fractions 207 

Permutations  and  Combinations 210 

SECTION  XVI. 

INVOLUTION   OF   BINOMIALS. 

Powers  of  a  Binomial 21 5 

Law  of  the  Exponents. — Coefficients 216 

Binomial  Theorem 219 

Theorem  applied  to  any  Polynomial 221 

When  the  Exponent  is  Negative 222 

When  the  Exponent  is  a  Fraction 224 

When  the  Exponent  is  a  Negative  Fraction 225 

Theorem  applied  to  find  the  Roots  of  Numbers 226 

SECTION  XVII. 

EVOLUTION   OF    POLYNOMIALS. 

Method  of  extracting  the  Square  Root  of  a  Polynomial 228 

Method  of  extracting  the  Square  Root  of  Numbers 229 

Method  of  extracting  the  Cube  Root  of  a  Polynomial 232 

1* 


Vlii  CONTENTS. 

Page 

Method  of  extracting  the  Cuhe  Root  of  Numbers 234 

Method  of  extracting  any  Boot  of  a  Polynomial 236 

Square  Root  of  a  ±Jb.— Formula 239 

SECTION  xvm. 

INFINITE   SERIES. 

Definitions.— Orders  of  Differences 243 

To  find  the  nth  Term  of  a  Series 244 

To  find  the  Sum  of  n  Terms  of  a  Series 246 

To  find  any  Term  of  a  Series  by  Interpolation 247 

Fractions  expanded  into  Infinite  Series. 249 

Infinite  Series  obtained  by  extracting  the  Square  Root 250 

Method  of  unknown  Coefficients  .—Rule 251 

SECTION  XIX. 

GENERAL   THEORY  OF  EQUATIONS. 

Definitions. — General  Form  of  E quations 255 

An  Equation  whose  Root  is  a  is  divisible  by  x  —  a 256 

An  Equation  of  the  mth  Degree  has  m  Roots 257 

Law  of  the  Coefficients  of  every  Equation 259 

What  Equations  have  no  Fractional  Roots 261 

How  the  Signs  of  all  the  Roots  may  be  changed 262 

The  Number  of  imaginary  Roots  must  be  even 263 

Number  of  Positive  and  Negative  Roots 264 

The  Limits  of  a  Root  determined 266 

The  Roots  may  be  increased  or  diminished  by  any  Quantity 267 

The  second  Term  of  an  Equation  taken  away 268 

Derived  Polynomials.— Equal  Roots 269 

Theorem  of  Sturm 272 

Elimination  by  means  of  the  Greatest  Common  Divisor 279 

Ctauul  wuli— -w«:«G  iuvuuij1*  J* ^J«ort> 
SECTION  XX. 

JOiil'*i  JjCBfliJiKjO 
SOLUTION   OF   NUMERICAL   EQUATIONS. 

To  find  the  Integral  Roots  of  an  Equation 281 

Homer's  Method  for  Incommensurable  Roots 284 

Equations  of  the  fourth  and  higher  Degrees 293 

Newton's  Method  of  Approximation 297 

Method  of  Double  Position 299 

The  different  Roots  of  Unity 301 

SECTION  XXI. 

LOGARITHMS. 

Logarithms. — Rule  for  the  Characteristic 304 

Multiplication  and  Division.— Involution  and  Evolution 306 

Computation  of  Logarithms 311 

Logarithmic  Series 313 

Naperian  Logarithms. — Common  Logarithms 316 

Exponential  Equations  solved 3.30 

Compound  Interest.— Increase  of  Population 321 

MISCELLANEOUS  EXAMPLES 305 


ALGEBRA. 

SECTION  I. 

PRELIMINARY  DEFINITIONS  AND  NOTATION. 

(Article  1.)  WHATEVER  is  capable  of  increase  or  diminution, 
or  will  admit  of  mensuration,  is  called  magnitude  or  quantity. 

A  sum  of  money,  therefore,  is  a  quantity,  since  we  may  in- 
crease it  or  diminish  it.  A  line,  a  surface,  a  weight,  and  other 
things  of  this  nature,  are  quantities  ;  but  an  idea  is  not  a  quantity. 

(2.)  Mathematics  is  the  science  of  quantity,  or  the  science 
which  investigates  the  means  of  measuring  quantity.  The 
operations  of  the  mind,  therefore,  such  as  memory,  imagina- 
tion, judgment,  &c.,  are  not  subjects  of  mathematical  investi- 
gation, since  they  are  not  quantities. 

(3.)  Mathematics  is  divided  into  pure  and  mixed.  Pure 
mathematics  comprehends  all  inquiries  into  the  relations  of 
magnitude  in  the  abstract,  and  without  reference  to  material 
bodies.  It  embraces  numerous  subdivisions,  such  as  Arith- 
metic, Algebra,  Geometry,  &c. 

In  the  mixed  mathematics  these  abstract  principles  are  ap- 
plied to  various  questions  which  occur  in  nature.  Thus,  in 
Surveying,  the  abstract  principles  of  Geometry  are  applied  to 
the  measurement  of  land  ;  in  Navigation,  the  same  principles 
are  applied  to  the  determination  of  a  ship's  place  at  sea ;  in 
Optics,  they  are  employed  to  investigate  the  properties  of 
light ;  and  in  Astronomy,  to  determine  the  distances  of  the 
heavenly  bodies. 

(4.)  Algebra  is  that  branch  of  mathematics  which  enables  us, 
by  means  of  letters  and  other  symbols,  to  abridge  and  generalize 
the  reasoning  employed  in  the  solution  of  all  questions  relating 
to  numbers. 

A. 


2  PRELIMINARY    DEFINITIONS    AND    NOTATION. 

Arithmetic  is  the  art  or  science  of  numbering.  It  treats  of 
the  nature  and  properties  of  numbers,  but  it  is  limited  to  cer- 
tain methods  of  calculation  which  occur  in  common  practice. 
Algebra  is  more  comprehensive,  and  has  been  called  by  New- 
ton, Universal  Arithmetic. 

(5.)  The  following  are  the  main  points  of  difference  between 
Arithmetic  and  Algebra. 

First,  the  operations  of  Algebra  are  more  general  than  those 
of  Arithmetic.  In  Arithmetic  we  represent  quantities  by  par- 
ticular numbers,  as  2,  5,  7,  &c.,  which  numbers  always  retain 
the  same  value.  The  results  obtained,  therefore,  are  applicable 
only  to  the  particular  question  proposed.  Thus,  if  it  is  re- 
quired to  find  the  interest  of  a  thousand  dollars  for  three 
months  at  six  per  cent.,  the  question  may  be  solved  by  Arith- 
metic, and  we  obtain  an  answer,  which  is  applicable  only  to 
this  problem. 

But  in  the  solution  of  a  general  Algebraic  problem  we  em- 
ploy letters,  to  which  any  value  may  be  attributed  at  pleasure. 
The  results  obtained,  therefore,  are  equally  applicable  to  all 
questions  of  a  particular  class.  Thus,  if  we  have  given  the 
sum  and  difference  of  two  quantities,  we  may  obtain  by  means 
of  Algebra  a  general  expression  for  the  quantities  themselves. 
This  result  will  always  be  found  true,  whatever  may  be  the 
magnitude  of  the  quantities.  Hence  Algebra  is  adapted  to  the 
investigation  of  general  principles,  while  Arithmetic  is  confined 
to  operations  upon  particular  numbers. 

Secondly,  Algebra  enables  us  to  solve  a  vast  number  of 
problems,  which  are  too  difficult  for  common  Arithmetic. 
Some  of  the  problems  in  Sections  VII.  and  VIII.  may  be 
solved  by  Arithmetical  methods ;  but  others  can  not  thus  be 
resolved,  particularly  such  problems  as  are  given  in  Sections 
XII.,  XIV.,  &c. 

Thirdly,  in  Arithmetic  all  the  different  quantities  which  en- 
ter into  a  problem  are  blended  together  in  the  result,  so  as  to 
leave  no  trace  of  the  operations  to  which  they  have  been  sub- 
jected. From  a  simple  inspection  of  the  result,  we  can  not 
tell  whether  it  was  derived  by  multiplication  or  division,  invo- 
lution or  evolution,  or  what  connection  it  has  with  the  given 
quantities  of  the  problem.  But  in  a  general  Algebraic  solu- 
tion, all  the  different  quantities  are  preserved  distinct  from  each 


PRELIMINARY    DEFINITIONS    AND    NOTATION.  3 

other,  and  we  see  at  a  glance  how  all  the  data  of  the  problem 
are  combined  in  the  result.  Illustrations  of  this  remark  will 
be  found  in  Section  VII.,  &c. 

Fourthly,  the  operations  of  Algebra  are  often  far  more  con- 
cise than  those  of  Arithmetic.  Thus,  although  some  of  the 
problems  in  Sections  VII.  and  VIII.  may  be  solved  Arith- 
metically, these  solutions  are  generally  much  more  tedious 
than  the  Algebraic.  This  advantage  which  is  possessed  by 
Algebra  is  partly  due  to  the  representation  of  the  unknown 
quantities  by  letters,  and  their  introduction  into  the  operations 
as  if  they  were  already  known,  and  partly  to  the  fact  that  the 
operations  of  multiplication,  division,  &c.,  are  at  first  merely 
indicated,  and  are  not  actually  performed  until  an  Algebraic 
expression  has  been  reduced  to  its  simplest  form. 

Finally,  perhaps  the  most  striking  difference  between  Arith- 
metic and  Algebra  springs  from  the  use  of  negative  quantities, 
which  give  rise  to  many  peculiar  results. 

The  full  purport  of  these  remarks  will  be  best  apprehended 
after  the  student  has  made  some  progress  in  the  study  of  Al- 
gebra. 

(6.)  A  definition  is  the  explanation  of  any  term  or  word.  It 
is  essential  to  a  perfect  definition  that  it  distinguish  the  thing 
defined  from  every  thing  else.  Thus,  if  we  say  that  man  is  a 
biped,  it  is  an  imperfect  definition  of  man,  because  there  are 
many  other  bipeds. 

(7.)  A  theorem  is  the  statement  of  some  property,  the  truth 
of  which  is  required  to  be  proved.  Thus  the  principle  that 
the  sum  of  the  three  angles  of  any  triangle  is  equal  to  two 
right  angles,  is  a  theorem,  the  truth  of  which  is  demonstrated 
by  Geometry. 

(8.)  A  problem  is  a  question  requiring  something  to  be  done. 
Thus,  to  draw  one  line  perpendicular  to  another  is  a  problem. 
Theorems  and  problems  are  both  known  by  the  general  term 
of  propositions. 

(9.)  A  determinate  problem  is  one  which  admits  of  a  certain 
or  definite  answer.  An  indeterminate  problem  commonly  ad- 
mits of  an  indefinite  number  of  solutions ;  although  when  the 
answers  are  required  in  positive  whole  numbers,  they  are  in 
some  cases  confined  within  certain  limits,  and  in  others  the 
problem  may  be  impossible. 


4  PRELIMINARY    DEFINITIONS    AND    NOTATION. 

(10.)  The  solution  of  a  problem  is  the  process  by  which  we 
obtain  the  answer  to  it.  A  numerical  solution  is  the  obtaining 
an  answer  in  numbers.  A  geometrical  solution  is  the  obtaining 
an  answer  by  the  principles  of  geometry.  A  mechanical  so- 
lution is  one  which  is  gained  by  trials. 

(11.)  The  principal  symbols  employed  in  Algebra  are  the 
following : 

The  sign  +  (an  erect  cross)  is  named  plus,  and  is  employed 
to  denote  the  addition  of  two  or  more  numbers.  Thus,  5+3 
signifies  that  we  must  add  3  to  the  number  5,  in  which  case 
the  result  is  8.  In  the  same  manner,  11+6  is  equal  to  17; 
14+10  is  equal  to  24,  &c. 

We  also  make  use  of  the  same  sign  to  connect  several  num- 
bers together.  Thus,  7+5+9  signifies  that  to  the  number  7 
we  must  add  5  and  also  9,  which  make  21. 

So,  also,  the  sum  of  8+5+13+11  +  1+3+10  is  equal  to  51. 

(12.)  In  order  to  generalize  numbers  we  represent  them  by 
letters,  as  a,  6,  c,  d,  &c.  Thus  the  expression  a -\-b  signifies 
the  sum  of  two  numbers,  which  we  represent  by  a  and  b,  and 
these  may  be  any  numbers  whatever.  In  the  same  manner, 
m+n+p+x  signifies  the  sum  of  the  numbers  represented  by 
these  four  letters.  If  we  knew,  therefore,  the  numbers  repre- 
sented by  the  letters,  we  could  easily  find  by  arithmetic  the 
value  of  such  expressions. 

The  first  letters  of  the  alphabet  are  commonly  used  to  rep- 
resent known  quantities,  and  the  last  letters  those  which  are 
unknown. 

(13.)  The  sign  —  (a  horizontal  line)  is  called  minus,  and  in- 
dicates that  one  quantity  is  to  be  subtracted  from  another. 
Thus,  8—5  signifies  that  the  number  5  is  to  be  taken  from  the 
number  8,  which  leaves  a .  remainder  of  3.  In  like  manner, 
12—7  is  equal  to  5,  and  20—14  is  equal  to  6,  &c. 

Sometimes  we  may  have  several  numbers  to  subtract  from 
a  single  one.  Thus,  16—5—4  signifies  that  5  is  to  be  subtract- 
ed from  16,  and  this  remainder  is  to  be  further  diminished  by 
4,  leaving  7  for  the  result.  In  the  same  manner,  50—1  —  3—5 
—7— 9  is  equal  to  25.  So,  also,  a— b  signifies  that  the  number 
designated  by  a  is  to  be  diminished  by  the  number  designated 
by  b. 

Quantities  preceded  by  the  sign  +  are  called  positive  quan 


PRELIMINARY    DEFINITIONS    AND    NOTATION.  5 

titles;  those  preceded  by  the  sign  — ,  negative  quantities. 
When  no  sign  is  prefixed  to  a  quantity,  +  is  to  be  understood. 
Thus,  a+b—c  is  the  same  as  +a+b— c. 

(14.)  The  sign  X  (an  inclined  cross)  is  employed  to  denote 
the  multiplication  of  two  or  more  numbers.  Thus,  3X5  signi- 
fies that  3  is  to  be  multiplied  by  5,  making  15.  In  like  man- 
ner, axb  signifies  a  multiplied  by  b ;  and  aXbXc  signifies  the 
continued  product  of  the  numbers  designated  by  «,  b,  and  c, 
and  so  on  for  any  number  of  quantities. 

Multiplication  is  also  frequently  indicated  by  placing  a  point 
between  the  successive  letters.  Thus,  a  .  b  .  c  .  d  signifies  the 
same  thing  as  aXbXcXd. 

Generally,  however,  when  numbers  are  represented  by  let- 
ters, their  multiplication  is  indicated  by  writing  them  in  suc- 
cession without  the  interposition  of  any  sign.  Thus,  a  b  sig- 
nifies the  same  thing  as  a .  b  or  a  X  b ;  and  a  b  c  d  is  equivalent 
to  a.  b.  c .  d,  or  aXbXcXd. 

It  must  be  remarked  that  the  notation  a .  b  or  a  b  is  seldom 
employed  except  when  the  numbers  are  designated  by  letters. 
If,  for  example,  we  attempt  to  represent  the  product  of  the 
numbers  5  and  6  in  this  manner,  5 . 6  might  be  confounded 
with  an  integer  followed  by  a  decimal  fraction ;  and  56  would 
be  read  fifty-six,  according  to  the  common  system  of  nota- 
tion. 

The  multiplication  of  numbers  may,  however,  be  expressed 
by  placing  a  point  between  them,  in  cases  where  no  ambiguity 
can  arise  from  the  use  of  this  symbol.  Thus,  1.2.3.4  is 
sometimes  used  to  represent  the  continued  product  of  the  num- 
bers 1,  2,  3,  4. 

(15.)  When  two  or  more  quantities  are  multiplied  together, 
each  of  them  is  called  a  factor.  Thus,  in  the  expression  7X5, 
7  is  a  factor,  and  so  is  5.  In  the  product  abc  there  are  three 
factors,  a,  fc,  c. 

When  a  quantity  is  represented  by  a  letter,  it  is  called  a 
literal  factor,  to  distinguish  it  from  a  numerical  factor,  which 
is  represented  by  an  Arabic  numeral.  Thus,  in  the  expression 
Sab,  5  is  a  numerical  factor,  while  a  and  b  are  literal  factors. 

(16.)  The  character  -r  (a  horizontal  line  with  a  point  above 
and  below)  shows  that  the  quantity  which  precedes  it  is  to  be 
divided  by  that  which  follows. 


6  PRELIMINARY    DEFINITIONS    AND    NOTATION. 

Thus,  24-T-6  signifies  that  24  is  to  be  divided  by  6,  making  4. 
So,  also,  a-^-b  is  a  divided  by  b. 

Generally,  however,  the  division  of  two  numbers  is  indi- 
cated by  writing  the  dividend  above  the  divisor,  and  drawing 
a  line  between  them. 

Thus,  24-7-6  and  a-^b  are  usually  written  ?  and  I. 

(17.)  The  sign  =  (two  horizontal  lines)  when  placed  be- 
tween two  quantities,  denotes  that  they  are  equal  to  each 
other. 

Thus,  7+6=13  signifies  that  the  sum  of  7  and  6  is  equal  to 
13.  So,  also,  $1  =  100  cents,  is  read  one  dollar  equals  one 
hundred  cents;  3  shillings =36  pence,  is  read  three  shillings 
are  equal  to  thirty-six  pence.  In  like  manner,  a=b  signifies 
that  a  is  equal  to  b;  and  a+b—c—d  signifies  that  the  sum  of 
the  numbers  designated  by  a  and  b  is  equal  to  the  difference 
of  the  numbers  designated  by  c  and  d. 

(18.)  The  symbol  >  is  called  the  sign  of  inequality ',  and 
when  placed  between  two  numbers,  denotes  that  one  of  them 
is  greater  than  the  other,  the  opening  of  the  sign  being  turned 
toward  the  greater  number. 

Thus,  3<5  signifies  that  3  is  less  than  5,  and  11>6  denotes 
that  11  is  greater  than  6.  So,  also,  a>6  shows  that  a  is 
greater  than  b,  and  c<c?  shows  that  c  is  less  than  d. 

(19.)  The  coefficient  of  a  quantity  is  the  number  or  letter 
prefixed  to  it,  showing  how  often  the  quantity  is  to  be  taken. 

Thus,  instead  of  writing  a+a+a+«+«,  which  represents 
five  a's  added  together,  we  write  5a,  where  5  is  the  coefficient 
of  a.  In  like  manner,  Wab  signifies  ten  times  the  product  of 
a  and  b.  The  coefficient  may  be  either  a  whole  number  or  a 
fraction.  Thus,  fa  signifies  three  fourths  of  a.  When  no  co- 
efficient is  expressed,  1  is  always- to  be  understood.  Thus,  la 
and  a  signify  the  same  thing. 

The  coefficient  may  be  a  letter  as  well  as  a  figure.  In  the 
expression  mx,  m  may  be  considered  as  the  coefficient  of  x, 
because  x  is  to  be  taken  as  many  times  as  there  are  units  in  m, 
If  m  stands  for  5,  then  mx  is  5  times  x. 

In  4ax,  4  may  be  considered  as  the  coefficient  of  ax,  or  4a 
may  be  considered  as  the  coefficient  of  x. 

(20.)  The  products  formed  by  the  successive  multiplication  of 
the  same  number  by  itself  are  called  the  powers  of  that  number. 


PRELIMINARY    DEFINITIONS    AND    NOTATION.  7 

Thus,  2X2=4,  the  second  power  of  2. 
2X2X2=8,  the  third  power. 
2X2X2X2=16,  the  fourth  power,  &c. 
So,  also,  3X3=9,  the  second  power  of  3. 

3X3X3=27,  the  third  power,  &c. 
Also,  aXa—aa,  the  second  power  of  a. 

aXaXa=aaa,  the  third  power,  &c. 

In  general,  any  power  of  a  quantity  is  designated  by  the 
number  of  factors  which  form  the  product. 

(21.)  For  the  sake  of  brevity,  powers  are  usually  expressed 
by  writing  the  root  once,  with  a  number  above  it  at  the  right 
hand,  showing  how  many  times  the  root  is  taken  as  a  factor. 
This  number  is  called  the  exponent  of  the  power. 
Thus,  instead  of 

art,  we  write  a2,  where  2  is  the  exponent  of  the  power. 
aaa,       "       a8,  where  3  is  the  exponent  of  the  power. 
aaaa,     "       a\  where  4  is  the  exponent  of  the  power. 
aaaaa,  "       «B,  where  5  is  the  exponent  of  the  power,  &c. 
When  no  exponent  is  expressed,  1  is  always  understood. 
Thus,  a1  and  a  signify  the  same  thing. 

Exponents  may  be  attached  to  figures  as  well  as  letters. 
Thus,  the  product  of  3  by  3  may  be  written  3a,  which  equals  9 
"  3X3X3  "  38,  "  27 

"  3X3X3X3  "  34,  "  81 

3X3X3X3X3         "  35,  "          243 

(22.)  A  root  of  a  quantity  is  a  factor,  which,  multiplied  by 
itself  a  certain  number  of  times,  will  produce  the  given  quan- 
tity. 

The  symbol  V  is  called  the  radical  sign,  and  when  pre- 
fixed to  a  quantity  denotes  that  its  root  is  to  be  extracted. 
Thus^ 

\/9,   or  simply  -v/9,  denotes  the  square  root  of  9,  which  is  3. 
V  64  denotes  the  cube  root  of  64,  which  is  4. 
vM6  denotes  the  fourth  root  of  16,  which  is  2. 
So,  also, 

v/a,  or  simply  ,/a,  is  the  square  root  of  a, 
ty 'a  denotes  the  third  or  cube  root  of  a. 
3/a  denotes  the  fourth  root  of  a. 

y  a  denotes  the  nth  root  of  a,  where  n  may  represent  any 
number  whatever. 


8  PRELIMINARY    DEFINITIONS    AND    NOTATION. 

The  number  placed  over  the  radical  sign  is  called  the  index 
of  the  root.  Thus,  2  is  the  index  of  the  square  root,  3  of  the 
cube  root,  4  of  the  fourth  root,  and  n  of  the  nth  root.  The  in- 
dex of  the  square  root  is  usually  omitted.  Thus,  instead  of 
Vab,  we  usually  write  Vab. 

(23.)  When  four  quantities  are  proportional,  the  proportion 
is  expressed  by  points,  as  in  arithmetic.  Thus,  a  :  b  : :  c  :  d, 
signifies  that  a  has  to  b  the  same  ratio  which  c  has  to  d. 

(24.)  A  vinculum ,  or  a  parenthesis  (    ),  indicates  that 

several  quantities  are  to  be  subjected  to  the  same  operation. 

Thus,  a+b+cXd,  or  (a+b+c)xd,  denotes  that  the  sum  of 
a,  b,  and  c  is  to  be  multiplied  by  d.  But  a+6+cXd,  denotes 
that  c  only  is  to  be  multiplied  by  d. 

When  the  parenthesis  is  used,  the  sign  of  multiplication  is 
generally  omitted.  Thus,  (a+b+c)Xd  is  the  same  as  (a+b 
+c)d,  or  d(a+b+c). 

(25.)  Three  dots  • .  •  are  sometimes  employed  to  denote 
therefore  or  consequently. 

A  few  other  symbols  are  employed  in  algebra,  in  addition  to 
those  here  enumerated,  which  will  be  explained  as  they  occur, 

(26.)  Every  number  written  in  algebraic  language,  that  is, 
by  aid  of  algebraic  symbols,  is  called  an  algebraic  quantity,  or 
an  algebraic  expression. 

Thus,  3a  is  the  algebraic  expression  for  three  times  the 
number  a. 

4a?  is  the  algebraic  expression  for  four  times  the  square  of 
the  number  a. 

7a864  is  the  algebraic  expression  for  seven  times  the  third 
power  of  a  multiplied  by  the  fourth  power  of  b. 

(27.)  An  algebraic  quantity,  not  composed  of  parts  which 
are  separated  from  each  other  by  the  sign  of  addition  or  sub- 
traction, is  called  a  monomial,  or  a  quantity  of  one  term,  or 
simply  a  term. 

Thus,  2a,  5bc,  and  7#y2  are  monomials. 

(28.)  An  algebraic  expression,  which  is  composed  of  several 
terms,  is  called  a  polynomial. 

Thus,  a+2b+5c—d  is  a  polynomial. 

A  polynomial  consisting  of  two  terms  only,  is  usually  called 
a  binomial ;  one  consisting  of  three  terms  is  called  a  trinomial. 

Thus,  3a+5fc  is  a  binomial,  and  a+3bc+zy  is  a  trinomial. 


PRELIMINARY    DEFINITIONS    AND    NOTATION.  9 

(29.)  The  numerical  value  of  an  algebraic  expression  is  the 
result  obtained  when  we  attribute  particular  values  to  the 
letters. 

Suppose  the  expression  is  2d*b. 

If  we  make  a=2  and  &=3,  the  value  of  this  expression  will 
be  2X2X2X3=24. 

If  we  make  a=4  and  b=3,  the  value  of  the  same  expression 
will  be  2X4X4X3=96. 

The  numerical  value  of  a  polynomial  is  not  affected  by 
changing  the  order  of  the  terms,  provided  we  preserve  their 
respective  signs. 

The  expressions  a*+2ab+bz,  a2+62+2a&,  b*+2ab+a\  have 
all  the  same  numerical  value. 

Thus,  if  a=5  and  &=2,  the  value  of  «2  will  be  25,  that  of 
2ab  will  be  20,  and  6a  will  be  4;  and  if  these  numbers  are 
added  together,  their  sum  will  be  the  same  in  whatever  order 
they  are  placed.  Thus, 

25  25  20  20  4  4 

20  4  25  4  25  20 

_4  ^0  _4  25  20  25 

49  49  49  49  49  49 

(30.)  Each  of  the  literal  factors  which  compose  a  term  is 
called  a  dimension  of  this  term ;  and  the  degree  of  a  term  is 
the  number  of  these  factors  or  dimensions.  A  numerical  co- 
efficient is  not  counted  as  a  dimension. 

Thus,  3a  is  a  term  of  one  dimension,  or  of  the  first  degree. 

Sab  is  a  term  of  two  dimensions,  or  of  the  second  degree. 

6a2ic8  is  a  term  of  six  dimensions,  or  of  the  sixth  degree. 

In  general,  the  degree,  or  the  number  of  dimensions  of  a 
term,  is  equal  to  the  sum  of  the  exponents  of  the  letters  con- 
tained in  the  term. 

Thus,  the  degree  of  the  term  5a6ac<f  is  1+2+1+3  or  7; 
that  is,  this  term  is  of  the  seventh  degree. 

(31.)  A  polynomial  is  said  to  be  homogeneous  when  all  its 
terms  are  of  the  same  degree. 

Thus,  2a— 3&+c,  is  of  the  first  degree  and  homogeneous. 

3aa— 4fl&+62,  is  of  the  second  degree  and  homogeneous. 
2a3+3a2c— 4cad,  is  of  the  third  degree  and  homogene- 
ous. 
5a3— 2ab+c,  is  not  homogeneous. 


10  PRELIMINARY    DEFINITIONS    AND    NOTATION. 

(32.)  Like  or  similar  terms  are  terms  composed  of  the  same 
letters  affected  with  the  same  exponents. 

Thus,  Sab  and  7ab  are  similar  terms. 

5#2c  and  3#2c   are  also  similar  terms. 

But  Sab1  and  Mb  are  not  similar ;  for,  although  they  contain 
the  same  letters,  the  same  letters  are  not  affected  with  the 
same  exponents. 

(33.)  The  reciprocal  of  a  quantity  is  the  quotient  arising 
from  dividing  a  unit  by  that  quantity. 

Thus,  the  reciprocal  of  2  is  | ;  the  reciprocal  of  a  is  ±. 

(34.)  A  few  examples  are  here  subjoined,  to  exercise  the 
learner  on  the  preceding  definitions  and  remarks. 

Examples  in  which  words  are  to  be  converted  into  algebraic 
symbols. 

Ex.  1.  What  is  the  algebraic  expression  for  the  following 
statement  ?  The  second  power  of  a,  increased  by  twice  the 
product  of  a  and  6,  diminished  by  c,  and  increased  by  d,  is 
equal  to  seventeen  times  /. 

Ans.  <z2+2«&-c+rf=17/. 

Ex.  2.  The  quotient  of  three  divided  by  the  sum  of  x  and 
four,  is  equal  to  twice  b  diminished  by  eight. 

Ex.  3.  One  third  of  the  difference  between  six  times  x  and 
four,  is  equal  to  the  quotient  of  five  divided  by  the  sum  of  a 
and  b. 

Ex.  4.  Three  quarters  of  x  increased  by  five,  is  equal  to 
three  sevenths  of  b  diminished  by  seventeen. 

Ex.  5.  One  ninth  of  the  sum  of  six  times  x  and  five,  added 
to  one  third  of  the  sum  of  twice  x  and  four,  is  equal  to  th^ 
product  of  a,  6,  and  c. 

Ex.  6.  The  quotient  arising  from  dividing  the  sum  of  a  and 
b  by  the  product  of  c  and  d,  is  equal  to  four  times  the  sum  of  e, 
/,  g,  and  h. 

(35.)  Examples  in  which  the  algebraic  signs  are  to  be  trans- 
lated into  common  language. 

,    x+a    x       d 

Ex.  1.  — r~+~=~ri:* 
b        c     a+b 

Ans.  The  quotient  arising  from  dividing  the  sum  of  x  and  a 
by  6,  increased  by  the  quotient  of  x  divided  by  c,  is  equal  to 
the  quotient  of  d  divided  by  the  sum  of  a  and  6. 

Ex.  2.  *la*+(b-c)X(d+e)=g+h. 


PRELIMINARY    DEFINITIONS    AND    NOTATION.  11 

How  should  the  preceding  example  be  read,  when  the  first 
parenthesis  is  omitted  ? 


3     -7+a 

_ 
Ex.  4. 


Ex.  5.  2a\/b*-ac=5(h+d+x). 

^-  6- 


(36.)  Find  the  value  of  the  following  expressions,  when  <z= 
6,  6=5,  and  c=4. 
Ex.  1.  a3+3<z&-ca. 

Ans.  36+90-16=110. 
Ex.  2.  atX(a-\-b)-2abc. 

Ans.  156. 


.  28. 


2bc 
Ex.  4.  C+-T= 


Ex.  6.  3^c+2aV2a+b+2c. 


Ex.  8.  In  the  expression    ,    _  3_  ,  let  «=3,  6=5,  c=2, 
and  2=6 ;  what  is  its  numerical  value? 


SECTION  II. 


ADDITION. 

(37  )  Addition  is  the  connecting  of  quantities  together  by 
means  of  their  proper  signs,  and  incorporating  such  as  can  be 
united  into  one  sum. 

It  is  convenient  to  distinguish  three  Cases. 

CASE  I. 

When  the  quantities  are  similar  and  have  the  same  signs. 

RULE. 

Add  the  coefficients  of  the  several  quantities  together,  and  to 
their  sum  annex  the  common  letter  or  letters,  prefixing  the  com- 
mon  sign. 

Thus,  the  sum  of  3a  and  5a  is  obviously  8a.  So,  also,  — 3a 
and  —  5a  make  —  8a;  for  the  minus  sign  before  each  of  the 
terms  shows  that  they  are  to  be  subtracted,  not  from  each 
other,  but  from  some  quantity  which  is  not  here  expressed ; 
and  if  3a  and  5a  are  to  be  successively  subtracted  from  the 
same  quantity,  it  is  the  same  as  subtracting  at  once  8a. 

EXAMPLES. 

2b+3x          a—2x* 
5b+7x        4a-3x* 
b+2x        3<z-5za 
4b+3x        7g—  x* 
a       - 

The  learner  must  continually  bear  in  mind  the  remark  of 
Art.  13,  that  when  no  sign  is  prefixed  to  a  quantity,  plus  is  al- 
ways to  be  understood. 


ADDITION.  13 


CASE  II. 

(38.)  When  the  quantities  are  similar,  but  have  different 
signs. 

RULE. 

Add  all  the  positive  coefficients  together,  and  also  all  those 
that  are  negative ;  subtract  the  least  of  these  results  from  the 
greater  ;  to  the  difference  annex  the  common  letter  or  letters,  and 
prefix  the  sign  of  the  greater  sum. 

Thus,  instead  of  7a— 4a,  we  may  write  3a,  since  these  two 
expressions  obviously  have  the  same  value. 

Also,  if  we  have  5a— 2a+3a— a,  this  signifies  that  from  5a 
we  are  to  subtract  2a,  add  3a  to  the  remainder,  and  then  sub- 
tract a  from  this  last  sum,  the  result  of  which  operation  is  5a. 
But  it  is  generally  most  convenient  to  take  the  sum  of  the  pos- 
itive quantities,  which  in  this  case  is  Sa ;  then  take  the  sum  of 
the  negative  quantities,  which  in  this  case  is  3a  ;  and  we  have 
Sa— 3a  or  5a,  the  same  result  as  before. 

EXAMPLES. 

— 3a        Gx+5ay  2ay—  7  —  2a*x  —  6a2+26 

—  3x+2ay  —  ay+  8  a?x  2a'2—3b 

x—6ay  2ay—  9  —3a?x  —  5ai—Sb 

—  a         2x+  ay  3ay—ll  7a?x  4d*—2b 


6x+2ay 


CASE  III. 
(39.)  When  some  of  the  quantities  are  dissimilar. 

RULE. 

Collect  all  the  like  quantities  together,  by  taking  their  sums  01 
differences  as  in  the  two  former  cases,  and  set  down  those  that 
are  unlike,  one  after  the  other,  with  their  proper  signs. 

Unlike  quantities  can  not  be  united  in  one  term.  Thus,  2a 
and  3b  neither  make  5a  nor  5b.  Their  sum  can  only  be  writ- 
ten 2a+3b. 


14  ADDITION. 


EXAMPLES. 

2xy—2x*  3xay-{-2ax        2ax^-220        2x    —I8y 

3x*  +  xy        —2xy*—  ax9       3x* —2ax        3xy  +10x 
x9—  xy        —3y*x+3ax*       5x*  —  3x          2x*y+25y 
4x* —3xy        —8x*y—  ax        3x  +100       12x*y—xy 
6#a—  xy  8x*—120 

(40.)  When  several  quantities  are  to  be  added  together,  it 
is  most  convenient  to  write  all  the  similar  terms  under  each 
other,  as  in  the  following  example. 
Ex.  1.  Add  together 

Ubc+4ad— 8ac+5cd 
8ac+1[bc—2ad+4mn 
2cd—3ab+5ac+  an 
9an — 2bc — 2ad+  5cd 
These  terms  may  be  written  thus : 

1 1  be + 4ad—  8ac + 5cd+ an + 4mn — 3ab 
7bc—2ad+8ac+2cd+9an 

— 2bc — 2ad+  5ac + 5cd 

Sum     16bc          +5ac+  12cd+  I0an+4mn—3ab. 
Ex.  2.  Add  together  the  quantities 
7m-\-3n  —I4p 
3a  +9n—  llm 
5p  —4m-}-  8n 
lln—2b—     m 

Ans.  3a—2b—9m+3ln—9p. 
Ex.  3.  Add  together 


27i-     ab*+5c3d 

*n-   5c3d+4mn*-8ab* 


9c*d  - 
Ans.  I 
Ex.  4.  Add  together 

3b-  a-Gc-115d-9f 

6c-5/—  d+     6e-3a 

3a-2b-3c+  27e-fll 

3e-7/+56-     8c+9d 

17c-6&-7<z-     2d-5e 

Ans.   - 


ADDITION.  15 

Ex.  5.  Add  together 


9b*x—   8hy*+10ky 
2ab2-3x*  -       Tfx-  4ky*-l5hy 
Sky-  hy*-22ac*-lOx*  —  4ab* 
9.r2  +  Qhy  +  2%2 
.  -9hy*+15ky—2ky*-9hy-4x\ 
(41.)  It  must  be  observed  that  the  term  addition  is  used  in 
a  more  extended  sense  in  algebra  than  in  arithmetic.     In  arith- 
metic, where  all  quantities  are  regarded  as  positive,  addition 
implies  augmentation.     The  sum  of  two  quantities  will  there- 
fore be  numerically  greater  than  either  quantity.     Thus  the 
sum  of  7  and  5  is  12,  which  is  numerically  greater  than  either 
5  or  7. 

But  in  algebra  we  consider  negative  as  well  as  positive 
quantities  ;  and  by  the  sum  of  two  quantities,  we  mean  their 
aggregate,  regard  being  paid  to  their  signs.  Thus  the  sum 
of  +7  and  —5  is  +2,  which  is  numerically  less  than  either  7 
or  5.  So,  also,  the  sum  of  +a  and  —b  is  a—  b.  In  this  case, 
the  algebraic  sum  is  numerically  the  difference  of  the  two 
quantities. 

This  is  one  instance,  among  many,  in  which  the  same  terms 
are  used  in  a  much  more  general  sense  in  the  higher  mathe- 
matics than  they  are  in  arithmetic. 

2 


SECTION  III. 


SUBTRACTION. 

(42.)  Subtraction  is  the  taking  of  one  quantity  from  anoth- 
er ;  or  it  is  finding  the  difference  between  two  quantities  or 
sets  of  quantities. 

Let  it  be  required  to  subtract  8—3  from  15. 

Now  8— 3  is  equal  to  5. 

And  5  subtracted  from  15  leaves  10. 

The  result,  then,  must  be  10.  But,  to  perform  the  operation 
on  the  numbers  as  they  were  given,  we  first  subtract  8  from 
15,  and  obtain  7.  This  result  is  too  small  by  3,  because  the 
number  8  is  larger  by  3  than  the  number  which  was  required 
to  be  subtracted.  Therefore,  in  order  to  correct  this  result, 
the  3  must  be  added,  and  we  have 

15-8+3=10,  as  before. 

Again,  let  it  be  required  to  subtract  c—d  from  a— b.  It  is 
plain,  that  if  the  part  c  were  alone  to  be  subtracted,  the  re- 
mainder would  be 

a— b— c. 

But  as  the.  quantity  actually  proposed  to  be  subtracted  is 
less  than  c  by  d,  too  much  has  been  taken  away  by  J,  and, 
therefore,  the  true  remainder  will  be  greater  than  a—b—c  by 
d,  and  will  hence  be  expressed  by 

a— b— c+d, 

where  the  signs  of  the  last  two  terms  are  both  contrary  to 
what  they  were  given  in  the  subtrahend. 

(43.)  Hence  we  deduce  the  following  general 

RULE. 
Conceive  the  signs  of  all  the  terms  of  the  subtrahend  to  be 


SUBTRACTION.  17 

changed  from  +  to  —  ,  or  from  —  to  +,  and  then  collect  the 
terms  together,  as  in  the  several  cases  of  addition. 

It  is  better  in  practice  to  leave  the  signs  of  the  subtrahend 
unchanged,  and  simply  conceive  them  to  be  changed  ;  that  is, 
treat  the  quantities  as  if  the  signs  were  changed  ;  for,  other- 
wise, when  we  come  to  revise  the  work  to  detect  any  error  in 
the  operation,  we  might  often  be  in  doubt  as  to  what  were  the 
signs  of  the  quantities  as  originally  proposed. 

EXAMPLES. 

From  5aa-26     6xy+Sx—2     10  —Sx—  3xy     4ax—2x*y 

Subtract       2a2+5fr     Sxy-Sx-l     -x+3  -  xy     3ax-5xy* 
Remainder  3a2—  Ib  7  —7x—2xy 

From  5a+4b-2c+7d        From       llxy+2y*-16x* 
Take  3a+2b+  c+5d         Take   --  4xy+6y*-18x* 
Remainder  2a+2b-3c+2d 

From       Gaby—  4xy+4xz          From  x*  +2xy-\-y* 
Take   -3aby+5xz+3xy          Take  x*-2xy+y'i 
Remainder     9aby—  xz—7xy 

From  3«2+  ax+2x*—  Ua^x+Wax*—  4x*    +5aV 
Take  2a'2-4ax+  xa—  15a2^  +  ll^2—  15aV—  4xa 


Subtraction  may  be  proved  as  in  Arithmetic,  by  adding  the 
remainder  to  the  subtrahend.  The  sum  should  be  equal  to  the 
minuend. 

(44.)  The  term  subtraction,  it  will  be  perceived,  is  used  in 
a  more  general  sense  in  algebra  than  in  arithmetic.  In  arith- 
metic, where  all  quantities  are  regarded  as  positive,  a  number 
is  always  diminished  by  subtraction.  But  in  algebra,  the  dif- 
ference between  two  quantities  may  be  numerically  greater 
than  either.  Thus,  the  difference  between  +a  and  —6  is  a+b. 

The  distinction  between  positive  and  negative  quantities 
may  be  illustrated  by  the  scale  of  a  thermometer.  The  de- 
grees above  zero  are  considered  positive,  and  those  below  zero 
negative.  From  five  degrees  above  zero  to  five  degrees  be- 
low zero,  the  numbers  stand  thus  : 

+  5,  +4,  +3,  +2,  +1,  0,  -1,  -2,  -3,  -4,  -5. 

The  difference  between  five  degrees  above  zero  and  five 
degrees  below  zero  is  ten  degree.s,  which  is  numerically  the 
sum  of  the  two  quantities. 

B 


18  SUBTRACTION. 

(45.)  In  practice,  it  is  often  sufficient  merely  to  indicate  the 
subtraction  of  a  polynomial,  without  actually  performing  the 
operation.  This  is  done  by  inclosing  the  polynomial  in  a  pa- 
renthesis, and  prefixing  the  sign  —  . 

Thus,  5a-3b+4c-(3a-2b+8c) 

signifies  that  the  entire  quantity  3a—  2b+8c  is  to  be  subtracted 
from  5a—  3fc+4c.  The  subtraction  is  here  merely  indicated. 
If  we  actually  perform  the  operation,  the  expression  becomes 

5a-3b+4c-3a+2b-8c 
or  2a  —  b—4c. 

(46.)  According  to  the  preceding  principle,  polynomials  may 
be  written  in  a  variety  of  forms. 

Thus,  a—  b—  c+d 
is  equivalent  to  a—  (b  +  c—d), 
or  to  a—  b—(c—d), 
or  to  a+  d—(b+c). 

Transformations  of  this  sort,  which  consist  in  decomposing 
a  polynomial  into  two  parts  separated  from  each  other  by  the 
sign  —  ,  are  of  frequent  use  in  algebra.  It  is  recommended  to 
the  student  to  write  out  polynomials  like  the  above,  contain- 
ing both  positive  and  negative  terms,  in  all  the  possible  modes, 
including  several  terms  in  a  parenthesis. 

In  the  following  examples,  let  the  results  all  be  reduced  to 
their  simplest  form. 

Ex.  1.  a+b-(2a-3b)-(5a+f!b)-(-13a+2b)=. 

Ex.  2.  37a—5f-(3a—2b—5c)  —  (6a—4b+3h)=. 

Ex.3.  8a*xy—5bx*y+l'7cxy*—9y6—(a'ixy+3bx'2y-13cxy*+ 


Ex.  4.  28ax*-16a*x*+25dix-13a*-(l8ax*+2Qa''x*-24a'ix 
—  7a4)=. 

(47.)  It  has  already  been  remarked,  in  Art.  5,  that  algebra 
differs  from  arithmetic  in  the  use  of  negative  quantities,  and  it 
js  important  that  the  beginner  should  obtain  clear  ideas  of  their 
nature. 

In  many  cases,  the  terms  positive  and  negative  are  merely 
relative.  They  indicate  some  sort  of  opposition  between  two 
classes  of  quantities,  such  that  if  one  class  should  be  added,  the 
other  ought  to  be  subtracted.  Thus,  if  a  ship  sails  alternately 
northward  and  southward,  and  the  motion  in  one  direction  is 


SUBTRACTION.  19 

called  positive,  the  motion  in  the  opposite  direction  should  be 
considered  negative. 

Suppose  a  ship,  setting  out  from  the  equator,  sails  north- 
ward 50  miles,  then  southward  27  miles,  then  northward  15 
miles,  then  southward  again  22  miles,  and  we  wish  to  deter- 
mine the  last  position  of  the  ship.  If  we  call  the  northerly 
motion  +,  the  whole  may  be  expressed  algebraically  thus : 

+  50-27+15-22, 

which  reduces  to  +16.  The  positive  sign  of  the  result  indi- 
cates that  the  ship  was  16  miles  north  of  the  equator. 

Suppose  the  same  ship  sails  again  8  miles  north,  then  35 
miles  south,  the  whole  may  be  expressed  thus : 

+  50-27+15-22+8-35, 

which  reduces  to  —11.  The  negative  sign  of  the  result  indi- 
cates that  the  ship  was  now  1 1  miles  south  of  the  equator. 

In  this  example  we  have  considered  the  northerly  motion  +, 
and  the  southerly  motion  —  ;  but  we  might  with  equal  pro- 
priety have  considered  the  southerly  motion  +,  and  the  north- 
erly motion  — .  It  is,  however,  indispensable  that  we  adhere 
to  the  same  system  throughout,  and  retain  the  proper  sign  of 
the  result,  as  this  sign  shows  whether  the  ship  was  at  any  time 
north  or  south  of  the  equator. 

In  the  same  manner,  if  we  consider  easterly  motion  +, 
westerly  motion  must  be  regarded  as  — ,  and  vice  versa. 
And  generally,  when  quantities  which  are  estimated  in  differ- 
ent directions  enter  into  the  same  algebraic  expression,  those 
which  are  measured  in  one  direction  being  treated  as  +,  those 
which  are  measured  in  the  opposite  direction  must  be  regard- 
ed as  — . 

So,  also,  in  estimating  a  man's  property,  gains  and  losses 
being  of  an  opposite  character,  must  be  affected  with  different 
signs.  Suppose  a  man,  with  a  property  of  1000  dollars,  loses 
300  dollars,  afterward  gains  100,  and  then  loses  again  400 
dollars,  the  whole  may  be  expressed  algebraically  thus : 

+  1000-300+100-400, 

which  reduces  to  +400.  The  +  sign  of  the  result  indicates 
that  he  has  now  400  dollars  remaining  in  his  possession.  Sup- 
pose he  further  gains  50  dollars  and  then  loses  700  dollars. 
The  whole  may  now  be  expressed  thus : 

+  1000-300  +  100-400  +  50-700, 


20  SUBTRACTION. 

which  reduces  to  —250.  The  —  sign  of  the  result  indicates 
that  his  losses  exceed  the  sum  of  all  his  gains  and  the  property 
originally  in  his  possession ;  in  other  words,  he  owes  250  dol- 
lars more  than  he  can  pay,  or,  in  common  language,  he  is  250 
dollars  worse  than  nothing. 

This  phraseology  must  not  be  regarded  as  wholly  figurative ; 
for,  in  algebra,  a  negative  quantity  standing  alone  is  regarded 
as  less  than  nothing;  and  of  two  negative  quantities,  that 
which  is  numerically  the  greatest  is  considered  as  the  least; 
for  if  from  the  same  number  we  subtract  successively  num- 
bers larger  and  larger,  the  remainders  must  continually  di- 
minish. Take  any  number,  5  for  example,  and  from  it  subtract 
successively  1,  2,  3,  4,  5,  6,  7,  8,  9,  &c.,  we  obtain 
5-1,  5-2,  5-3,  5-4,  5-5,  5-6,  5-7,  5-8,  5-9,  &c.,  or 
reducing 

4,3,2,  1,0,  -1,  -2,  -3,  -4. 

Whence  we  see  that  —1  should  be  regarded  as  smaller  than 
nothing;  —2  less  than  —  1 ;  —  3  less  than  —2, 


SECTION  IV. 


MULTIPLICATION. 

(48.)  Multiplication  is  repeating  the  multiplicand  as  many 
times  as  there  are  units  in  the  multiplier. 

When  several  quantities  are  to  be  multiplied  together,  the 
result  will  be  the  same  in  whatever  order  the  multiplication  is 
performed. 

This  may  be  demonstrated  in  the  following  manner : 

Let  unity  be  repeated  five  times  upon  a  horizontal  line,  and 
let  there  be  formed  four  such  parallel  lines. 


Then  it  is  plain  that  the  number  of  units  in  the  table  is  equal 
to  the  five  units  of  the  horizontal  line,  repeated  as  many  times 
as  there  are  units  in  a  vertical  column ;  that  is,  to  the  product 
of  5  by  4.  But  this  sum  is  also  equal  to  the  four  units  of  a 
vertical  line  repeated  as  many  times  as  there  are  units  in  a 
horizontal  line  ;  that  is,  to  the  product  of  4  by  5.  Therefore, 
the  product  of  5  by  4  is  equal  to  the  product  of  4  by  5.  For 
the  same  reason,  2X3X4  is  equal  to  2X4X3,  or  4X3X2,  or 
3X4X2,  the  product  in  each  case  being  24.  So,  also,  if  a,  b, 
and  c  represent  any  three  numbers,  we  shall  have  dbc  equal  to 
bca  or  cab. 

It  is  convenient  to  consider  the  subject  of  multiplication  un- 
der three  Cases. 


22  MULTIPLICATION. 


CASE  I. 

(49.)  When  both  the  factors  are  monomials. 

From  Article  14,  it  appears  that,  in  order  to  represent  the 
multiplication  of  two  monomials,  such  as  Sabc  and  5def,  we 
may  wrke  these  quantities  in  succession  without  interposing 
any  sign,  and  we  shall  have 

SabcSdef. 

But,  according  to  the  principle  stated  in  the  preceding  ai- 
ticle,  this  result  may  be  written 

3X5abcdef,  or  I5abcdef. 
Hence  we  deduce  the  following 

RULE. 

Multiply  the  coefficients  of  the  two  terms  together,  and  to  the 
product  annex  all  the  different  letters  in  succession. 

EXAMPLES. 

Multiply  I2a        5a        *ldb        7axy        Gxyz 
By  Sb         6x         Sac         Gay  ayz 

Product   BQab 

From  Article  48,  it  appears  to  be  immaterial  in  what  order 
the  letters  of  a  term  are  arranged ;  it  is,  however,  generally 
most  convenient  to  arrange  them  alphabetically. 

(50.)  We  have  seen  in  Art.  21,  that  when  the  same  letter 
appears  several  times  as  a  factor  in  a  product,  this  is  briefly 
expressed  by  means  of  an  exponent.  Thus,  aaa  is  written  «3, 
the  number  3  showing  that  a  enters  three  times  as  a  factor. 
Hence,  if  the  same  letters  are  found  in  two  monomials  which 
are  to  be  multiplied  together,  the  expression  for  the  product 
may  be  abbreviated  by  adding  the  exponents  of  the  same  let- 
ters. Thus,  if  we  are  to  multiply  a3  by  a2,  we  find  a8  equiva- 
lent to  aaa,  and  a2  to  aa:  Therefore  the  product  will  be  aaaaa, 
which  may  be  written  a5,  a  result  which  we  might  have  ob- 
tained at  once  by  adding  together  3  and  2,  the  exponents  of 
the  common  letter  a. 

Hence,  since  every  factor  of  both  multiplier  and  multipli- 
cand must  appear  in  the  product,  we  have  the  following 


MULTIPLICATION.  23 


RULE  FOR  THE  EXPONENTS. 

Powers  of  the  same  quantity  may  be  multiplied  by  adding 
heir  exponents. 

EXAMPLES. 

Multiply      8«2fc2          2ez362c         5aW  2aW 

By  7abcd*         Sabc*          7a*b*c3d         5a*bc* 

Product     56as£V<f 

CASE  II. 

(51.)   When  the  multiplicand  is  a  polynomial. 

If  a-\-b  is  to  be  multiplied  by  c,  this  implies  that  the  sum  of 
the  units  in  a  and  b  is  to  be  repeated  c  times  ;  that  is,  the  units 
in  b  repeated  c  times  must  be  added  to  the  units  in  a  repeated 
also  c  times.  Hence  we  deduce  the  following 

RULE. 

Multiply  each  term  of  the  multiplicand  separately  by  the  mul- 
tiplier, and  add  together  the  products. 

EXAMPLES. 

Multiply  3a+2b       a?+2x  +  l        3ya  +  5o:y+2       3x*+xy+2y* 
By  4a  4x  xy  Sx^y 

Product  12a*+8ab 

CASE  III. 

(52.)  When  both  the  factors  are  polynomials. 

If  a+b  is  to  be  multiplied  by  c+d,  this  implies  that  the 
quantity  a+b  is  to  be  repeated  as  many  times  as  there  are 
units  in  the  sum  of  c  and  d;  that  is,  we  are  to  multiply  a-\-b 
by  c  and  d  successively,  and  add  the  partial  products.  Hence 
we  deduce  the  following 

RULE. 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier  separately,  and  add  together  the  products. 

2* 


24  MULTIPLICATION. 

EXAMPLES. 

Multiply     a+b  3x+2y  ax+b  30+  x 

2x+3y  cx+d  2a+4x 


Product  aa 

When  several  terms  in  the  product  are  similar,  it  is  most 
convenient  to  set  them  under  each  other,  and  then  unite  them 
by  the  rules  for  addition. 

(53.)  The  examples  thus  far  given  in  multiplication  have 
been  confined  to  positive  quantities,  and  the  products  have  all 
been  positive.  We  must  now  establish  a  general  rule  for  the 
signs  of  the  product. 

First,  if  +a  is  to  be  multiplied  by  +6,  this  signifies  that  +a 
is  to  be  repeated  as  many  times  as  there  are  units  in  6,  and 
the  result  is  +ab.  That  is,  a  plus  quantity  multiplied  by  a 
plus  quantity  gives  a  plus  result. 

Secondly,  if  —a  is  to  be  multiplied  by  +6,  this  signifies  that 
—a  is  to  be  repeated  as  many  times  as  there  are  units  in  b. 
Now  —a  taken  twice  is  obviously  —  2a,  taken  three  times  is 
—  3a,  &c.  ;  hence,  if  —a  is  repeated  b  times,  it  will  make  —  ba 
or  —  ab.  That  is,  a  minus  quantity  multiplied  by  a  plus  quan- 
tity gives  minus. 

Thirdly,  to  determine  the  sign  of  the  product  when  the  mul- 
tiplier is  a  minus  quantity,  let  it  be  proposed  to  multiply  8—5 
by  6—2.  By  this  we  understand  that  the  quantity  8—5  is  to 
be  repeated  as  many  times  as  there  are  units  in  6—2.  If  we 
multiply  8—5  by  6,  we  obtain  48—30;  that  is,  we  have  re- 
peated 8—5  six  times.  But  it  was  only  required  to  repeat  the 
multiplicand  four  times,  or  (6—2).  We  must  therefore  dimin- 
ish this  product  by  twice  (8—5),,  which  is  16—10;  and  this 
subtraction  is  performed  by  changing  the  signs  of  the  subtra- 
hend ;  hence  we  have 

48-30-16+10, 

which  is  equal  to  12.  This  result  is  obviously  correct;  for 
8—  5  is  equal  to  3,  and  6—  2  is  equal  to  4;  that  is,  it  was  re- 
quired to  multiply  3  by  4,  the  result  of  which  is  12,  as  found 
above. 

In  order  to  generalize  this  reasoning,  let  it  be  proposed  to 
multiply  a—  b  by  c—  d. 

If  we  multiply  a—  b  by  c,  we  obtain  ac—bc.     But  a—b 


iMULTIPLICATlON.  25 

only  to  be  taken  c—  d  times;  therefore,  in  this  first  operation, 
we  have  repeated  it  too  many  times  by  the  quantity  d.  Hence, 
to  have  the  true  product,  we  must  subtract  d  times  a—  b  from 
ac—bc.  But  d  times  a—  b  is  equal  to  ad—bd,  which,  subtract- 
ed from  ac—bc,  gives 

ac—bc—  ad-}-  bd. 

Thus  we  see  that  +a  multiplied  by  —  d  gives  —ad;  and  —  b 
multiplied  by  —  d  gives  -\-bd.  Hence  a  plus  quantity  multi- 
plied by  a  minus  quantity  gives  minus,  and  a  minus  quantity 
multiplied  by  a  minus  quantity  gives  plus. 

(54.)  The  preceding  results  may  be  briefly  expressed  as  fol- 
lows : 

4-  multiplied  by  +,  and  —  multiplied  by  —  ,  give  +. 

+  multiplied  by  —  ,  and  —  multiplied  by  +,  give  —  . 

Or,  the  product  of  two  quantities  having  the  same  sign,  has 
the  sign  plus  ;  the  product  of  two  quantities  having  different 
signs,  has  the  sign  minus. 

(55.)  The  whole  doctrine  of  multiplication  is  therefore  com- 
prehended in  the  following 

RULE. 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier,  and  add  together  all  the  partial  products,  observing 
that  like  signs  require  +  in  the  product,  and  unlike  signs  —  . 

EXAMPLE    I. 

Multiply  5a4-  2asb+  4a*b* 

By  a3-  4a26+  2b* 

r  5a7-  2a'b+  4a*b* 

.Partial      J  _20a°&+  8a6b*-16a'bs 
Pr°ductSl 


_ 
Result  5a7-22a*b+l2a*b*-  6a4&3-4a364+8a2&5. 

Ex.  2.  Multiply  4a3-5a26-8a&2+2&3  by  2a*-3ab-4b\ 

Ans.  8a5  -  22a*b  -  1  7a3&2  +  48a263  +  26ab*  -  Sb6. 
Ex.  3.  Multiply  3a2—  5bd+ef  by  —5a*+4bd-8ef. 

Ans.  —  15a4+37«26c?-29a2e/-20ra2+44^e/-8e3/a. 
Ex.  4.  Multiply  x*+2x3+3x*+2x+l  by  af-gar+l. 

Ans.  xe—2x3+I. 
Ex.  5.  Multiply  14a3e-6aafo+ca  by  14a3e+6a2fo-cs. 


26  MULTIPLICATION. 

Ex.  6.  Multiply  3a3+35a86-17afe2-13&3  by  3aa+26a&- 
57fca. 

(56.)  Since  in  the  multiplication  of  two  monomials  every 
factor  of  both  quantities  appears  in  the  product,  it  is  obvious 
that  the  degree  of  the  product  will  be  equal  to  the  sum  of  the 
degrees  of  the  multiplier  and  multiplicand.  Hence,  also,  if 
two  polynomials  are  homogeneous,  their  product  will  be  homo- 
geneous. 

Thus*,,  in  the  first  of  the  preceding  examples,  all  the  terms 
of  the  multiplicand  being  of  the  fourth  degree,  and  those  ol 
the  multiplier  of  the  third  degree,  all  the  terms  of  the  product 
are  of  the  seventh  degree.  For  a  like  reason,  in  the  second 
example,  all  the  terms  of  the  product  are  of  the  fifth  degree ; 
In  the  third  example,  they  are  of  the  fourth  degree ;  and  in  the 
sixth  example,  they  are  of  the  fifth  degree. 

This  remark  will  enable  us  to  detect  any  error  in  the  mul- 
tiplication, so  far  as  concerns  the  exponents.  For  example,  if 
we  find  in  one  of  the  terms  of  a  product  which  should  be  ho- 
mogeneous, the  sum  of  the  exponents  equal  to  6,  while  in  all 
the  other  terms  it  is  equal  to  7,  a  mistake  has  evidently  been 
committed  in  the  formation  of  one  of  the  terms. 

(57.)  When  the  product  arising  from  the  multiplication  of 
two  polynomials  does  not  admit  of  any  reduction  of  similar 
terms,  the  whole  number  of  terms  in  the  product  is  equal  to  the 
number  of  terms  in  the  multiplicand,  multiplied  by  the  number 
of  terms  in  the  multiplier. 

Thus,  if  we  have  five  terms  in  the  multiplicand  and  four 
terms  in  the  multiplier,  the  whole  number  of  terms  in  the  prod- 
uct will  be  5X4,  or  20.  In  general,  if  there  be  m  terms  in  the 
multiplicand  and  n  terms  in  the  multiplier,  the  whole  number 
of  terms  in  the  product  will  be  mXn. 

(58.)  If  the  product  contains  similar  terms,  the  number  of 
terms  in  the  product  when  reduced  may  be  much  less ;  but  it 
is  important  to  observe,  that  among  the  different  terms  of  the 
product  there  are  always  two  which  can  not  be  combined  with 
any  others.  These  are, 

1.  The  term  arising  from  the  multiplication  of  the  two  terms 
affected  with  the  highest  exponent  of  the  same  letter. 

2.  The  term  arising  from  the  multiplication  of  the  two  terms 
fleeted  with  the  lowest  exponent  of  the  same  letter. 


?,i  L;  LTI  i  •  L  i  c  A  T  i  o  x .  27 

For  it  is  evident,  from  the  rule  of  exponents,  that  these  two 
partial  products  must  involve  the  letter  in  question,  the  one 
with  a  higher,  and  the  other  with  a  lower  exponent  than  any 
of  the  other  partial  products,  and  therefore  can  not  be  similar 
to  any  of  them.  Hence  the  product  of  two  polynomials  can 
never  contain  less  than  two  terms. 

(59.)  For  many  purposes,  it  is  sufficient  merely  to  indicate 
the  multiplication  of  two  polynomials,  without  actually  per- 
forming the  operation.  This  is  effected  by  inclosing  the  quan- 
tities in  parentheses,  and  writing  them  in  succession  with  or 
without  the  interposition  of  any  sign. 

Thus,  (a+b+c)  (d+e+f)  signifies  that  the  sum  of  a,  b,  and 
c  is  to  be  multiplied  by  the  sum  of  d,  e,  and  /. 

When  the  multiplication  is  actually  performed,  the  expres- 
sion is  said  to  be  expanded. 

(60.)  The  following  Theorems  are  of  such  extensive  appli- 
cation that  they  should  be  carefully  committed  to  memory. 

THEOREM  I. 

The  square  of  the  sum  of  two  quantities  is  equal  to  the  square 
of  the  first,  plus  twice  the  product  of  the  first  by  the  second,  plus 
the  square  of  the  second. 

Thus,  if  we  multiply  a  +  b 

By  a+  b 

«a+  ab 

ab+V 
We  obtain  the  product  a~+2ab+ff. 

Hence,  if  we  wish  to  obtain  the  square  of  a  binomial,  we 
can  write  out  the  terms  of  the  result  at  once  according  to  this 
theorem  without  the  necessity  of  performing  an  actual  multi- 
plication.' 

EXAMPLES. 

2.  (a+3by=.  7.  (5a8+by=. 

3.  (3a+3by=.  8. 

4.  (4a+3by=.  9. 

5.  (5a*+by=.  10. 

This  theorem  deserves  particular  attention,  for  one  of  the 


28  MULTIPLICATION. 

most  common  mistakes  of  beginners  is  to  call  the  square  of 
a+b  equal  to  <za+6a. 

THEOREM  II. 

(61.)  The  square  of  the  difference  of  two  quantities  is  equal 
to  the  square  of  the  first,  minus  twice  the  product  of  the  first  and 
second,  plus  the  square  of  the  second. 

Thus,  if  we  multiply  a  —  b 

By  a-  b 

a*-  ab 

-  ab+b* 
We  obtain  the  product  d*—2ab+b*. 

EXAMPLES. 

1.  (a-2b)*=.  6.  (7aa-6)a=. 

2.  (2a-36)2=.  7. 

3.  (5<z-46)3=.  8. 

4.  (6a*-xy=.  9.  (2-i)a:=. 

5.  (6aa-3x)a=.  10.  (4-i)a=. 

Here,  also,  beginners  often  commit  the  mistake  of  putting 
the  square  of  a— b  equal  to  aa— b\ 

THEOREM  III. 

(62.)  The  product  of  the  sum  and  difference  of  two  quantities 
is  equal  to  the  difference  of  their  squares. 

Thus,  if  we  multiply  a  +b 

By  a  -b 

a*+ab 

-ab-b* 
We  obtain  the  product  a3— 6a. 

EXAMPLES. 

1.  (2a+b)  (2a-b)=. 

2.  (Za+4b)  (3a-4b)=. 

3.  (la+x)  C?a-x)=. 

4.  (7ab+x)  (7ab-x)  =  . 

5.  (Sa+b)  (8a-b)=. 

6.  (Qa+lbc)  (8a-7bc)  = 

7.  (5 

8.  5 


MULTIPLICATION.  29 

9.  (3+})  (3-1)=. 
10.  (4+i)  (4-1)-. 

The  student  should  be  drilled  upon  examples  like  those  ap- 
pended to  the  preceding  theorems  until  he  can  produce  the  re- 
sults mentally  with  as  great  facility  as  he  could  read  them  if 
exhibited  upon  paper. 

The  utility  of  these  theorems  will  be  the  more  apparent,  the 
more  complicated  the  expressions  to  which  they  are  applied. 
Frequent  examples  of  their  application  will  be  seen  hereafter. 

(63.)  The  same  theorems  will  enable  us  to  resolve  many 
complicated  expressions  into  their  factors. 

I.  Resolve  a2+4a&+4&2  into  its  factors. 

Ans.  (a+2b)  (a+2b). 

2.  Resolve  «2— 6ab+9b*  into  its  factors. 

3.  Resolve  9a2— 24«&+16&2  into  its  factors. 

4.  Resolve  a*  —  V  into  three  factors. 

5.  Resolve  ae—b°  into  its  factors. 

6.  Resolve  a9— bs  into  four  factors. 

7.  Resolve  25at-60a2b3+3Gb6  into  its  factors. 

8.  Resolve  ri*+2n+l  into  its  factors. 

9.  Resolve  4wi2?i2—- 4mn-\-l  into  its  factors. 

10.  Resolve  49a4&4-168a363+144a2&2  into  its  factors. 

II.  Resolve  7i3+2n2+7i  into  three  factors. 

12.  Resolve  1  —  ^  into  two  factors. 

13.  Resolve  4— into  two  factors. 


MULTIPLICATION  BY  DETACHED  COEFFICIENTS. 

(64.)  The  coefficients  of  a  product  depend  simply  upon  the 
coefficients  of  the  two  factors,  and  not  upon  the  literal  parts 
of  the  terms.  Hence  we  may  obtain  the  coefficients  of  the 
product  by  multiplying  the  coefficients  of  the  multiplicand  sev- 
erally by  the  coefficients  of  the  multiplier.  To  these  coeffi- 
cients the  proper  letters  may  afterward  be  annexed.  This 
will  be  best  understood  from  a  few  examples. 

Thus,  take  the  first  example  of  Art.  52,  to  multiply  a+b  by 
a+b. 


30  MULTIPLICATION. 

The  coefficients  of  the  multiplicand  are     1  +  1 
"  ".       multiplier  1  +  1 

+ 

Coefficients  of  the  product  1+2+1 

or,  supplying  the  letters,  we  obtain  a2+2a&+62, 

which  is  the  same  result  as  before  ol  tained. 

Ex.  2.  Multiply  3a*+4ax—  5#2  by  2a*-6ax+4x\ 
Coefficients  of  multiplicand  3+  4—  5 

"  multiplier  2—6+4 

6+  8-10 
-18-24+30 

+  12+16-20 
Coefficients  of  the  product  6—10—22+46—20 

It  may  seem  difficult  in  this  case  to  supply  the  letters  ;  but 
a  little  consideration  will  render  it  perfectly  plain.  Thus, 
3a2X2a2  is  equal  to  6a4  ;  hence  a4  is  the  proper  letter  to  be  at- 
tached to  the  first  coefficient.  For  the  same  reason,  x*  is  the 
proper  letter  to  be  attached  to  the  last  coefficient.  Moreover, 
we  see  that  both  the  proposed  polynomials  are  homogeneous, 
and  of  the  second  degree.  Hence  the  product  must  be  ho- 
mogeneous, and  of  the  fourth  degree.  The  powers  of  a  must 
decrease  successively  by  unity,  beginning  with  the  first  term, 
while  those  of  x  increase  by  unity.  Hence  the  required  prod- 
uct is 


Ex.  3.  Multiply  x*+x*y+xy*+y*  by  x-y. 

Ex.4.  Multiply  x'—  3z2+3#-l  by  x*—2z+l. 

Ex.  5.  Multiply  2a3—  3«62+563  by  2a-5b. 

If  we  should  proceed  with  this  example  precisely  in  the  same 
manner  as  with  the  preceding,  we  should  commit  an  error  by 
attempting  to  unite  terms  which  are  dissimilar.  The  reason 
is,  that  the  multiplicand  does  not  contain  the  usual  complete 
series  of  powers  of  a.  The  term  containing  the  second  power 
of  a  is  wanting.  This  does  not  render  the  method  inapplica- 
ble, but  it  is  necessary  to  preserve  dissimilar  terms  distinct 
from  each  other  ;  and  since,  while  we  are  are  operating  on  the 
coefficients,  we  have  not  the  advantage  of  the  letters  to  indi- 
cate what  are  similar  terms,  we  supply  the  place  of  the  defi 


MULTIPLICATION.  31 

cient  term  by  a  cipher.      The  operation  will  then  proceed 
with  entire  regularity. 

2+   0-3+   5 

2-   5 

4+   0-6+10 
— 10-0+15— 25 

4-10-6+25-25 
Hence  the  product  is 


Ex.  6.  Multiply  2a3-3ab*+5b3  by  2a2-563. 

Here  there  is  a  term  in  each  polynomial  to  be  supplied  by  a 
cipher. 

The  preceding  examples  are  intended  to  lead  the  student  to 
consider  the  properties  of  coefficients  by  themselves,  and  pre- 
pare him  for  some  investigations  which  are  to  follow,  particu- 
larly in  Section  XX.  The  beginner,  however,  in  attempting 
to  apply  the  method,  must  be  cautious  not  to  unite  dissimilar 
terms. 


SECTION  V. 


DIVISION. 

(65.)  The  object  of  division  in  Algebra  is  the  same  as  in 
Arithmetic,  viz.,  The  product  of  two  factors  being  given,  and 
one  of  the  factors,  to  find  the  other  factor. 

The  dividend  is  the  product  of  the  divisor  and  quotient,  the 
divisor  is  the  given  factor,  and  the  quotient  is  the  factor  re- 
quired to  be  found. 

CASE  I. 

(66.)  When  the  divisor  and  dividend  are  both  monomials. 

Suppose  we  have  63  to  be  divided  by  7.  We  must  find  such 
a  factor  as,  multiplied  by  7,  will  give  exactly  63.  We  per- 
ceive that  9  is  such  a  number,  and  therefore  9  is  the  quotient 
obtained  when  we  divide  63  by  7. 

Also,  if  we  have  to  divide  ab  by  #,  it  is  evident  that  the 
quotient  will  be  b  ;  for  a  multiplied  by  b  gives  the  dividend  ab. 
So,  also,  I2mn  divided  by  3m  gives  4n ;  for  3m  multiplied  by 
4n  makes  I2mn. 

Suppose  we  have  a6  to  be  divided  by  a2.  We  must  find  a 
number  which,  multiplied  by  a2,. will  produce  a&.  We  perceive 
that  a8  is  such  a  number ;  for,  according  to  Art.  50,  we  multi- 
ply a9  by  a\  by  adding  the  exponents  2  and  3,  making  5. 
That  is,  the  exponent  3  of  the  quotient  is  found  by  subtracting 
2,  the  exponent  of  the  divisor,  from  5,  the  exponent  of  the  divi- 
dend. Hence  the  following 

RULE  OF  EXPONENTS  IN  DIVISION. 

In  order  to  divide  quantities  expressed  by  different  powers 
of  the  same  letter,  subtract  the  exponent  of  the  divisor  from  the 
exponent  of  the  dividend. 


DIVISION. 


33 


EXAMPLES. 

Divide      a6          a?         b6         c9          W          x°          ym 

Quotient  a" 

Let  it  be  required  to  divide  35a6  by  5a8.  We  must  find  a 
quantity  which,  multiplied  by  5a\  will  produce  35a6.  Such  a 
quantity  is  7a3 ;  for,  according  to  Arts.  49  and  50,  7#3  X5a2  is 
equal  to  35a\  Therefore,  35a5  divided  by  5a2  gives  for  a 
quotient  7a3 ;  that  is,  we  have  divided  35,  the  coefficient  of  the 
dividend,  by  5,  the  coefficient  of  the  divisor,  and  have  sub- 
tracted the  exponent  of  the  divisor  from  the  exponent  of  the 
dividend. 

(67.)  Hence,  for  the  division  of  monomials,  we  have  the  fol- 
lowing 

RULE. 

1 .  Divide  the  coefficient  of  the  dividend  by  the  coefficient  of  the 
divisor. 

2.  Subtract  the  exponent  of  each  letter  in  the  divisor  from  the 
exponent  of  the  same  letter  in  the  dividend. 

EXAMPLES 

1.  Divide  20x*  by  4x.  Ans.  5za. 

2.  Divide  25a3xy*  by  5ay*. 

3.  Divide  72«6V  by  I2b3x. 

4.  Divide  77aW  by  Ilab3c\ 

5.  Divide  272a3b*c*x°  by  lla^cx*. 

6.  Divide  250x7y*z3  by  5xyz3. 

7.  Divide  48«3&Vd  by  12ab~c. 

8.  Divide  150a66W3  by  3Qa*b*d*. 

(68.)  The  rule  given  in  Art.  66  conducts,  in  some  cases,  to 
negative  exponents. 

Thus,  let  it  be  required  to  divide  a3  by  a5.  We  are  directed 
to  subtract  the  exponent  of  the  divisor  from  the  exponent  of 
the  dividend.  We  thus  obtain 


But  a3  divided  by  a6  may  be  written  — ;  and  since  the  value 


34  DIVISION. 

of  a  fraction  is  not  altered  by  dividing  both  numerator  and  de- 
nominator by  the  same  quantity,  this  expression  is  equivalent 

to—. 
a3 

Hence  <z~2  is  the  same  as  — , 

and  these  expressions  may  be  used  indifferently  for  each  other. 
So,  also,  if  aa  is  to  be  divided  by  a\  this  may  be  written 


In  the  same  manner  we  find 

--=1. 

a 

That  is,  the  reciprocal  of  a  quantity  is  equal  to  the  same  quan- 
tity with  the  sign  of  its  exponent  changed. 

So,  also,  rr=—    =ab~*c~l. 

DC        C 

ad~*      a 

And  —  =—•  =r^. 

b       bd 

(69.)  Hence  any  factor  may  be  transferred  from  the  numer- 
ator to  the  denominator  of  a  fraction,  or  from  the  denominator 
to  the  numerator,  by  changing  the  sign  of  its  exponent. 

Thus,  T 


That  is,  the  denominator  of  a  fraction  may  be  entirely  re- 
moved, and  an  integral  form  be  given  to  any  fractional  ex- 
pression. 

This  use  of  negative  exponents  must  be  understood  simply 
as  a  convenient  notation,  and  not  as  a  method  of  actually  de- 
stroying the  denominator  of  a  fraction.  Still  this  new  nota 
tion  has  many  advantages,  and  is  often  employed,  as  will  be 
seen  hereafter. 

When  the  division  can  not  be  exactly  performed,  it  may  be 
expressed  in  the  form  of  a  fraction,  and  this  fraction  may  be 


DIVISION.  35 

reduced  to  its  lowest  terms,  according  to  a  method  to  be  ex- 
plained in  Art.  S3. 

(70.)  It  frequently  happens  that  the  exponents  of  certain  let- 
ters in  the  dividend  are  the  same  as  in  the  divisor. 

Let  it  be  required  to  divide  a2  by  a2.  The  quotient  is  ob- 
viously 1,  for  every  number  is  contained  in  itself  once.  But 
if  we  apply  the  rule  of  exponents,  Art.  66,  we  shall  have 

«2~~2  or  a°. 
Hence  a°=l. 

Again,  let  it  be  required  to  divide  am  by  am.  Tne  quotient 
is  obviously  1,  as  before ;  and  applying  the  rule  of  exponents, 

we  obtain 

am— m  or  ao 

That  is,  every  quantity  affected  with  the  exponent  zero,  is  equal 
to  unity. 

This  notation  has  the  advantage  of  preserving  a  trace  of  a 
letter  which  has  disappeared  in  the  operation  of  division. 
Thus,  let  it  be  required  to  divide  a362  by  <2262.  The  quotient 
will  be  ab°.  This  expression  is  of  the  same  value  as  a  alone, 
and  is  commonly  so  written.  If,  however,  it  was  important  to 
indicate  that  the  -letter  b  originally  entered  into  the  expression, 
this  might  te  done  without  at  ail  affecting  the  value  of  the  re- 
sult by  writing  it 

ab°. 

(71.)  The  proper  sign  to  be  prefixed  to  a  quotient  is  readily 
deduced  from  the  principles  already  established  for  multipli- 
cation. The  product  of  the  divisor  and  quotient  must  be  equal 
to  the  dividend.  Hence, 

because  +aX+b=  +ab^\  (+ab—+b=+a. 

-ax+l >=-ab  ^  therefore  ^  -ab-+b=-a. 

+aX—b=—ab  j  j  —ab b=+a. 

—  aX-b=+abJ  [+ab b=  —  a. 

Hence  we  have  the  following 

RULE  FOB  THE  SIGNS. 

When  both  the  dividend  and  divisor  have  the  same  sign,  the 
quotient  will  have  the  sign  +  ;  when  they  have  different  signs, 
the  quotient  will  have  the  sign  —. 


36  DIVISION. 

X 


EXAMPLES. 

Ex.  1.  Divide  —  15ay8  by  Say. 
Ex.  2.  Divide  —  18ax*y  by  —  9ax. 
Ex.  S.  Divide  I5Qa*bc  by  —  5ac. 
Ex.  4.  Divide  40a36V  by  —  abc. 

CASE  II. 

(72.)  When  the  divisor  is  a  monomial,  and  the  dividend  a 
polynomial. 

We  have  seen,  Art.  51,  that  when  a  single  term  is  multi- 
plied into  a  polynomial,  the  former  enters  into  every  term  of 
the  latter. 

Thus,  a(a+b)=a*+ab. 

Hence  (a*+ab)-t-a=a+b. 

Whence  we  deduce  the  following 

RULE. 

Divide  each  term  of  the  dividend  by  the  divisor,  as  in  the  for- 
mer case. 

\ 

EXAMPLES. 

Ex.  1.  Divide  3x*+6x*+3ax-I5x  by  Sx. 

Ans.  x*+2x+a— 5. 

Ex.  2.  Divide  3abc+12abx-9a'ib  by  Sab. 

Ex.  S.  Divide  4008&3+60a2&2-17a&  by  -ab. 

Ex.  4.  Divide  I5a*bc—  10flcx2+5ac2^2  by  — 5#2c. 

Ex.  5.  Divide  6aVy6-12«8;ry+15ffV?/3  by  3«V?/a. 

Ex.  6.  Divide  xn+i— xn+z+xn-^-xn+4  by  x". 

Ex.  7.  Divide  12«y— 16«y+20«y-28ay  by  — 4«y. 

CASE  III. 

(73.)  When  the  divisor  and  dividend  are  both  polynomials. 

Let  it  be  required  to  divide  2afe+«2+62  by  a+b. 

The  object  of  this  operation  is  to  find  a  third  polynomial 
which,  multiplied  by  the  second,  will  reproduce  the  first. 

It  is  evident  that  the  dividend  is  composed  of  all  the  partial 
products  arising  from  the  multiplication  of  each  term  of  the 
divisor  by  each  term  of  the  quotient,  these  products  being  add- 
ed together  and  reduced.  Hence,  if  we  can  discover  a  term 


DIVISION.  37 

of  the  dividend  which  is  derived  without  reduction  from  the 
multiplication  of  a  term  of  the  divisor  by  a  term  of  the  quo- 
tient, then  dividing  this  term  by  the  corresponding  term  of  the 
divisor,  we  shall  be  sure  to  obtain  a  term  of  the  quotient. 

But  from  Art.  58,  it  appears  that  the  term  a2,  which  contains 
the  highest  exponent  of  the  letter  #,  is  derived,  without  reduc- 
tion, from  the  multiplication  of  the  two  terms  of  the  divisor 
and  quotient  which  are  affected  with  the*  highest  exponent  of 
the  same  letter.  Dividing  then  the  term  a2  by  the  term  a  of 
the  divisor,  we  obtain  a,  which  we  are  certain  must  be  one 
term  of  the  quotient  sought.  Multiplying  each  term  of  the  di- 
visor by  a,  and  subtracting  this  product  from  the  proposed 
dividend,  the  remainder  may  be  regarded  as  the  product  of 
the  divisor  by  the  remaining  terms  of  the  quotient.  We  shall 
then  obtain  another  term  of  the  quotient  by  dividing  that  term 
of  the  remainder  affected  with  the  highest  exponent  of  a.  by 
the  term  a  of  the  divisor,  and  so  on. 

Thus  we  perceive  that  at  each  step  we  are  obliged  to  search 
for  that  term  of  the  dividend  which  is  affected  with  the  high- 
est exponent  of  one  of  the  letters,  and  divide  it  by  that  term 
of  the  divisor  which  is  affected  with  the  highest  exponent  of 
the  same  letter.  We  may  avoid  the  necessity  of  searching  for 
this  term  by  arranging  the  terms  of  the  divisor  and  dividend 
in  the  order  of  the  powers  of  one  of  the  letters. 

The  operation  will  then  proceed  as  follows : 


The  arranged  dividend  =«Q 


ab 


a+b=  the  divisor. 


a+b=  the  quotient. 
2=  first  remainder. 
ab+b* 


It  is  generally  convenient  in  Algebra  to  place  the  divisor  on 
the  right  of  the  dividend,  and  the  quotient  directly  under  the 
divisor. 

(74.)  From  this  investigation  we  deduce  the  following 

RULE  FOE  THE  DIVISION  OF  POLYNOMIALS. 

1  .  Arrange  the  dividend  and  divisor  according  to  the  powers 
of  the  same  letter. 


38  DIVISION. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  the  result  will  be  the  first  term  of  the  quotient. 

3.  Multiply  the  divisor  by  this  term,  and  subtract  the  product 
rom  the  dividend. 

4.  Divide  the  first  term  of  the  remainder  by  the  first  term  of 
he  divisor,  the  result  will  be  the  second  term  of  the  quotient. 

5.  Multiply  the  divisor  by  this  term,  and  subtract  the  product 
from  the  last  remainder.     Continue  the  same  operation  till  all 
the  terms  of  the  dividend  are  exhausted. 

If  the  divisor  is  not  exactly  contained  in  the  dividend,  the 
quantity  which  remains  after  the  division  is  finished  must  be 
placed  over  the  divisor  in  the  form  of  a  fraction,  and  annexed 
to  the  quotient. 

EXAMPLES. 

1.  Divide  2a*b+b3+2ab*+a3  by  a*+b*+ab. 

Ans.  a+b. 

2.  Divide  x3— a*+3a*x— Sax*  by  x—a. 

Ans.  x*—2ax+a9. 

3.  Divide  «6+ze+2aV  by  a'-az+z*. 

Ans.  a4+a3z+az5+z*. 

4.  Divide  a*-I6a3x*+G4x°  by  a*-4ax+4x\ 

5.  Divide  a*+6d1x*—4a3x+x*—4ax3  by  a*— 2ax+x\ 

Ans.  a?—2ax+x*. 

6.  Divide  x*+x*y* +?/4  by  x*+xy+y*. 

7.  Divide  I2x*-192  by  3x-6. 

Ans.  4x3+8x*+16x+32. 

8.  Divide  6x6-6y*  by  2x*-2y*. 

9.  Divide  <z6+3a2&4-3a4&2-66  by  a3-3a*b+3ab*-b3. 

Ans.  a3+3a*b+3ab*+b3. 

10.  Divide  a3— b3  by  a— b. 

11.  Divide  «4-64  by  a— b. 

If  the  first  term  of  the  arranged  dividend  is  not  divisible  by 
the  first  term  of  the  arranged  divisor,  the  complete  division  is 
impossible. 

(75.)  Hitherto  we  have  supposed  the  terms  of  the  quotient 
to  be  obtained  by  dividing  that  term  of  the  dividend  affected 
with  the  highest  exponent  of  a  certain  letter.  But,  from  the 
second  remark  of  Art.  58,  it  appears  that  the  term  of  the  divi- 
dend affected  with  the  lowest  exponent  of  any  letter  is  derived, 


DIVISION.  3t 

without  reduction,  from  the  multiplication  of  a  term  of  the  di- 
visor by  a  term  of  the  quotient.  Hence  we  may  obtain  a  term 
of  the  quotient  by  dividing  the  term  of  the  dividend  affected 
with  the  lowest  exponent  of  any  letter,  by  the  term  of  the  di- 
visor containing  the  lowest  power  of  the  same  letter,  and 
nothing  prevents  our  operating  upon  the  highest  and  lowest 
exponents  of  a  certain  letter  alternately  in  the  same  example. 

(76.)  From  the  examples  of  Art.  74,  we  perceive  that  a*—b* 
is  divisible  by  a—b ;  and  a*—b*  is  divisible  by  a—b.  We  shall 
find  the  same  to  hold  true,  whatever  may  be  the  value  of  the 
exponents  of  the  two  letters.  That  is,  the  difference  of  any 
two  powers  of  the  same  degree  is  divisible  by  the  difference  of 
their  roots. 

Thus,  let  us  divide  ab— b5  by  a—b. 
a*—b5   a—b 


The  first  term  of  the  quotient  is  <z4,  and  the  first  remainder 
is  a*b—b5,  which  may  be  written 

b(a*-V). 

Now  if,  after  a  division  has  been  partially  performed,  the  re- 
mainder is  divisible  by  the  divisor,  it  is  obvious  that  the  divi- 
dend is  completely  divisible  by  the  divisor.  But  we  have  al- 
ready found  that  a4— b4  is  divisible  by  a—b;  therefore  a6— b* 
is  also  divisible  by  a—b ;  and  in  the  same  manner  it  may  be 
proved  that  a6— b6  is  divisible  by  a—b,  and  so  on. 

To  exhibit  this  reasoning  in  a  more  general  form,  let  us 
represent  any  exponent  whatever  by  the  letter  n,  and  let  us 
divide  an—bn  by  a—b. 


an—bn 
an-ban~l 


a-b 


First  remainder  =  ba"—{—bn. 

Dividing  an  by  a,  we  have,  by  the  rule  of  exponents,  a"—1  for 
the  quotient.  Multiplying  a—b  by  this  quantity,  and  subtract- 
ing the  product  from  the  dividend,  we  have  for  the  first  re- 
mainder ban~l  —  bn,  which  may  ^e  written 

b(an-l-b"-^). 

Now  if  this  remainder  is  divisible  by  a—b,  it  is  obvious  that 
the  dividend  is  divisible  by  a—b.  That  is  to  say,  if  the  differ- 

3 


40  DIVISION. 

§ 

ence  of  the  same  powers  of  two  quantities  is  divisible  by  their 
difference,  the  difference  of  the  powers  of  the  next  higher  de- 
gree is  also  divisible  by  that  difference.  Therefore,  since  a*—b* 
is  divisible  by  a—  b,  a*—  I?  must  be  divisible  by  a—  b  ;  also, 
a6—  b\  and  so  on. 

The  quotients  obtained  by  dividing  the  difference  of  the 
powers  of  two  quantities  by  the  difference  of  those  quantities, 
follow  a  simple  law.     Thus, 
(a*-b*)-r-(a-b)=a+b. 
(a*-b*)+(a-b)=a*+ab+b't. 


(a*-b6)+(a-b)=a*+a*b+a*b*+ab9+b*. 

&c.,  &c. 


The  exponents  of  a  decrease  by  unity,  while  those  of  b  in- 
crease by  unity. 

(77.)  It  may  also  be  proved  that  the  difference  of  two  even 
powers  of  the  same  degree  is  divisible  by  the  sum  of  their  roots. 
Thus/ 

(a*-b*)-(a+b)=a-b. 
(a4-b4)-(a+b)=a*-a*b+ab*-b\ 
(a°-b')-(a+b)=a*-atb+a9b*-a*b9+ab'-b*. 
&c.,         &c.,  &c. 

Also,  the  sum  of  two  odd  powers  of  the  same  degree  is  divisi- 
ble by  the  sum  of  their  roo/s 
Thus, 


&c.,         &c., 

(78.)  The  preceding  principles  will  enable  us  to  resolve  va- 
rious algebraic  expressions  into  their  factors. 

1.  Resolve  a9—  63  into  its  factors. 

Ans.  (a*+ab+b*)  (a-b). 

2.  Resolve  «3+63  into  its  factors. 

3.  Resolve  a6—  b*  into  four  factors. 

4.  Resolve  a8—  863  into  its  factors. 

5.  Resolve  8«3—  1  into  its  factors. 

6.  Resolve  8«3—  8b*  into  three  factors. 


DIVISION.  41 

7.  Resolve  1+27&3  into  its  factors. 

8.  Resolve  8aa+27ba  into  its  factors. 

(79.)  One  polynomial  can  not  be  divided  by  another  poly- 
nomial containing  a  letter  which  is  not  found  in  the  dividend ; 
for  it  is  impossible  that  one  quantity  multiplied  by  another 
which  contains  a  certain  letter,  should  give  a  product  not  con- 
taining that  letter. 

A  monomial  is  never  divisible  by  a  polynomial,  because 
every  polynomial  multiplied  by  another  quantity  gives  a  prod- 
uct containing  at  least  two  terms  not  susceptible  of  reduction. 

Yet  a  binomial  may  be  divided  by  a  polynomial  containing 
any  number  of  terms. 

Thus,  a4— 54  is  divisible  by  <z8+a26+a&2+&3,  and  gives  for  a 
quotient  a— b. 

So,  also,  a  binomial  may  be  divided  by  a  polynomial  of  a 
hundred  terms,  a  thousand  terms,  or,  indeed,  any  finite  num- 
ber. 

DIVISION  BY  DETACHED  COEFFICIENTS. 

(80.)  We  have  shown,  in  Art.  64,  how  multiplication  may 
sometimes  be  conveniently  performed  by  operating  upon  the 
coefficients  alone.  The  same  principle  is  applicable  to  divi- 
sion. Thus,  take  the  example  of  Art.  73,  to  divide  «2 
by  a+fe;  we  may  proceed  as  follows  : 


1+2  +  1 
1  +  1 

1  +  1 

1  +  1 

1+1 
1+1 

The  coefficients  of  the  quotient  are  1  +  1.  Moreover,  a*+a 
=a ;  and  therefore  a  is  the  first  term  of  the  quotient,  and  b  the 
second. 

Ex.  2.  Divide  a;4— 3ax3—8aix^l8a3x— Qatbyx*+2ax—2a* 


1  —  3-   8  +  18-8 
1+2-  2 

1+2-2 

1-5+4 

-5-  6+18-8 
-5-10+10 

4+   8-8 
4+  8-8 
The  coefficients  of  the  quotient  are  1  —  5+4,  and  it  remains  to 


42  DIVISION. 

supply  the  letters.  Now  x*+x*=x*-,  and  «4-r-«a=aa.  Hence 
x9,  ax,  and  a2  are  the  literal  parts  of  the  terms,  and  therefore 
the  quotient  is 


Ex.  3.  Divide  604—  96  by  30-6. 

Here,  as  we  have  the  fourth  power  of  a  without  the  lower 
powers,  we  must  supply  the  coefficients  of  the  absent  terms, 
as  in  multiplication,  with  zero. 


6+  0+0+0-96 
6-12 


3-6 


2+4+8  +  16 


12 
12-24 


24 

24-48 
~48-96 
48-96 
But  a*+a=a* ;  hence  the  quotient  is 

203+4aa+8a+16. 
Ex.  4.  Divide  3y+3xya— 4x*y— 4x*  by  x+y. 

Ans.  3y-4za. 
Ex.  5.  Divide  8«5— 4a*x— 2a*x'+a*x*  by  4«a-xa. 

Ans.  2a*—d*x. 

Ex.  6.   Divide  «8+4a6-8«4-25a3+35aa+21a-28  by  «a+ 
50+4. 

Ans.  «4-03-7«a+140-7. 


SECTION  VI. 


FRACTIONS. 

(81.)  When  a  quotient  is  expressed  as  described  in  Art.  16, 
by  placing  the  divisor  under  the  dividend  with  a  line  between 
them,  it  is  called  a  fraction  ;  the  dividend  is  called  the  numer- 
ator, and  the  divisor  the  denominator  of  the  fraction.  Alge- 
braic fractions  do  not  differ  essentially  from  arithmetical  frac- 
tions, and  the  same  principles  are  applicable  to  both. 

The  following  principles  form  the  basis  of  most  of  the  oper- 
ations upon  fractions : 

1.  In  order  to  multiply  a  fraction  by  any  number,  we  must 
multiply  its  numerator,  or  divide  its  denominator  by  that  num- 
ber. 

Thus,  the  value  of  the  fraction  —  is  b.     If  we  multiply  the 

numerator  by  a,  we  obtain  —  or  ab ;  and  if  we  divide  the  de- 
nominator of  the  same  fraction  by  «,  we  obtain  also  ab  ;  that 
is,  the  original  value  of  the  fraction  b  has  been  multiplied  by  a. 

2.  In  order  to  divide  a  fraction  by  any  number,  we  must  di- 
vide its  numerator  or  multiply  its  denominator  by  that  number. 

27 

Thus,  the  value  of  the  fraction  —  is  ab.     If  we  divide  the 

a 

numerator  by  a,  we  obtain  —  or  b ;  and  if  we  multiply  the  de- 
nominator of  the  same  fraction  by  a,  we  obtain  ^  or  b  ;  that 

CL 

is,  the  original  value  of  the  fraction  ab  has  been  divided  by  a. 

3.  The  value  of  a  fraction  is  not  changed  if  we  multiply  or 
divide  both  numerator  and  denominator  by  the  same  number. 


44  FRACTIONS. 

db    abx    abxy 

Thus,  — = = ^=6. 

a      ax      axy 

Every  quantity  which  is  not  expressed  under  a  fractional 
form,  is  called  an  entire  quantity. 

An  algebraic  expression  composed  partly  of  an  entire  quan- 
tity and  partly  of  a  fraction,  is  called  a  mixed  quantity. 

(82.)  The  proper  sign  to  be  prefixed  to  a  fraction  may  be 
determined  by  the  rules  already  established  for  division.  The 
sign  prefixed  to  the  numerator  of  a  fraction  affects  merely  the 
dividend;  the  sign  prefixed  to  the  denominator  affects  merely 
the  divisor ;  but  the  sign  prefixed  to  the  dividing  line  of  a 
fraction  affects  the  quotient. 

Thus,  — =+b,  for  4-  divided  by  +  gives  +. 
— =— &,  for  —  divided  by  +  gives  — . 
_•—•=—&,  for  +  divided  by  —  gives  — . 
— =+&,  for  —  divided  by  —  gives  +. 

So,  also, =—b,  for  this  shows  that  the  former  quotient 

b  is  to  be  subtracted,  which  is  done  by  changing  its  sign. 

=  +&,  because  the  former  quotient  —6  is  to  be 

subtracted,  whence  it  becomes  +b. 

=  +b,  for  the  same  reason ; 

— a 

and =— b,  also  for  the  same  reason. 

—a 

Hence  we  have  the  following  equivalent  forms : 

ab     —ah        —ab         ab 

— — = = =  -f  &  - 

a      —a  a         —a 

—ab     ab         ab         —ab 

also,  = — = = =  — o. 

a       —a         a          —a 

That  is,  of  the  three  signs  belonging  to  the  numerator,  de- 
nominator, and  dividing  line  of  a  fraction,  any  two  may  be 
changed  from  +  to  —  or  from  —  to  +,  without  affecting  the 
value  of  the  fraction. 


FRACTIONS.  45 

In  the  examples  of  fractions  here  employed  for  illustration, 
both  numerator  and  denominator  have  consisted  of  monomials. 
The  same  principles  are  applicable  to  polynomials ;  but  it 
must  be  remarked,  that  by  the  sign  of  the  numerator  we  un- 
derstand the  entire  numerator  as  distinguished  from  the  sign 
of  any  one  of  its  terms  taken  singly. 

a+b+c .  —a—b—c 

Thus, —  is  equal  to  H 1 . 

a  a 

When  no  sign  is  prefixed  either  to  the  terms  of  a  fraction  or 
to  its  dividing  line,  4-  is  always  to  be  understood. 

EEDUCTION  OF  FRACTIONS. 

PROBLEM  I. 
(83.)  To  reduce  a  fraction  to  lower  terms. 

RULE. 

Divide  both  numerator  and  denominator  by  any  quantity 
which  will  divide  them  both  without  a  remainder. 

According  to  Remark  3,  Art.  81,  this  will  not  change  the 
value  of  the  fraction. 

£•4 

Also,  aM==^r  (dividing  both  numerator  and  de- 

5a  b     DO^ 

nominator  by  d*b.) 

ax*         ax 

And  — . 

ax+x      a+x 

If  the  numerator  and  denominator  are  both  divided  by  their 
greatest  common  divisor,  it  is  evident  the  fraction  will  be  re- 
duced to  its  lowest  terms.  The  method  of  finding  the  greatest 
common  divisor  is  considered  in  Section  XV. ;  but  in  the  fol- 
lowing examples  the  greatest  common  divisor  is  easily  found, 
by  resolving  the  quantities  into  factors  according  to  methods 
already  indicated. 

EXAMPLES. 

cx-\-x* 

1.  Reduce  -3— ;— 3-  to  its  lowest  terms. 
*> 

Ans.  ~ 


46  FRACTIONS. 


14<za— lab 

2.  Reduce  — rr-  to  its  lowest  terms. 

Wac—5bc 


x*—  a* 
3.  Reduce    4_  4  to  its  lowest  terms. 


2x*—16x-C> 

4.  Reduce  ^ — — ~  to  its  lowest  terms. 

— 


la 

Ans.  — . 
5c 


Ans.  •        a. 


5- Reduce  to  its  lowest  terms- 


6.  Reduce  —„ — ft  ,  ,  ,  a  to  its  lowest  terms. 

aa— 2ab+b* 

a*—x* 

7.  Reduce  -5 — ; — 5  to  its  lowest  terms. 

a*— 2ax+x* 


PROBLEM  II. 
(84.)  To  reduce  a  fraction  to  an  entire  or  mixed  quantity. 

RULE. 

Divide  the  numerator  by  the  denominator  for  the  entire  part, 
and  place  the  remainder,  if  any,  over  the  denominator  for  the 
fractional  part. 

27 
Thus,  — =27-r-5=5f. 

Also,  =(ax+a>)-^-x=a-{ — . 

X  X 

EXAMPLES. 

1.  Reduce to  an  entire  quantity. 

•K 


FRACTIONS.  47 


2.  Reduce  — — r —  to  a  mixed  quantity. 

fla+a;a 

3.  Reduce to  a  mixed  quantity. 

a—x 


Ans. 




4.  Reduce —  to  an  entire  quantity. 

x—y 

5.  Reduce to  a  mixed  quantity. 

DX 


6.  Reduce rr to  a  mixed  quantity. 

PROBLEM  III. 

(85.)  To  reduce  a  mixed  quantity  to  the  form  of  a  fraction 

RULE. 

Multiply  the  entire  part  by  the  denominator  of  the  fraction ; 
to  the  product  add  the  numerator  with  its  proper  sign,  and  place 
the  result  over  the  denominator. 

3X5+2     15+2     17 
Thus,  S|=  —      — =T- 

This  result  may  be  proved  by  the  preceding  Rule.     For 


_ 
c          c  c 


EXAMPLES. 

aa— #2 

1.  Reduce  x-\ to  the  form  of  a  fraction. 

x 


2.  Reduce  x-\ — - —  to  the  form  of  a  fraction. 
2a 


Ans.  — . 
x 


3ax+x* 


48  FRACTIONS. 


_ 

3.  Reduce  5H  —  -  —  to  the  form  of  a  fraction. 

4.  Reduce  1  H  --  to  the  form  of  a  fraction. 


_ 

5.  Reduce  1  +2x-\  —  -  —  to  the  form  of  a  fraction. 

. 

6.  Reduce  7H  —  r~n~  to  the  form  of  a  fraction. 


PROBLEM  IV. 
(86.)  To  reduce  fractions  to  a  common  denominator. 

RULE. 

Multiply  each  numerator  into  all  the  denominators,  except  its 
own,  for  a  new  numerator,  and  all  the  denominators  together  for 
a  common  denominator. 

EXAMPLES. 

a         c 

1.  Reduce  7  and  -3  to  a  common  denominator. 

b         a 

ad     be 
bd;  bd" 

Here  it  will  be  seen  that  the  numerator  and  denominator  of 
the  first  fraction  are  both  multiplied  by  d,  and  in  the  second 
fraction  they  are  both  multiplied  by  b.  The  value  of  the  frac- 
tions, therefore,  is  not  changed  by  this  operation. 

2.  Reduce  j-  and to  equivalent  fractions  having  a  com- 

*  mon  denominator. 

ac     ab+b* 


Ans. 


be'      be    ' 


3x  2& 
3.  Reduce  —  ,  —  ,  and  d  to  fractions  having  a  common  de- 


nominator. 

O 


4.  Reduce  -,  —  ,  and  a+-j-  to  fractions  having  a  common 
denominator. 


FRACTIONS.  49 

5.  Reduce  -,  —  ,  and  -  to  fractions  having  a  common 

2    7  a—  x 

denominator. 

np      '•>*  ^|  _.  1  1  i        Q* 

6.  Reduce  -,  —  —  -,  and  —  —  to  fractions  having  a  common 

O          O  1  ~T~X 

denominator. 

7.  Reduce  —  -  and  -  to  fractions  having  a  common  de- 

3x3  4z 

nominator. 

Following  the  Rule,  we  obtain 
Sax 


which  fractions  have  a  common  denominator,  and  are  equiva- 
lent to  those  originally  proposed.  Nevertheless,  it  may  be  ob- 
served, that  these  fractions  are  not  reduced  to  their  least  com- 
mon denominator,  for  every  term  is  divisible  by  x.  The  least 
common  denominator  is  the  least  common  multiple  of  the  de- 
nominators of  the  proposed  fractions. 

A  common  multiple  of  two  or  more  numbers  is  any  number 
which  they  will  divide  without  a  remainder  ;  and  the  least  com- 
mon multiple  is  the  least  number  which  they  will  so  divide. 
Thus,  12x*  is  the  least  common  multiple  of  3x*  and  4x  ;  and 
the  above  fractions  reduced  to  their  least  common  denomina- 
tor are 

8a         , 
15  and 


The  least  common  multiple  of  two  numbers  is  their  product 
divided  by  their  greatest  common  divisor. 

o  c 

8.  Reduce  —  and  —  to  equivalent  fractions  having  the  least 

common  denominator. 

The  product  of  the  denominators  is  294,  which,  divided  by 
7  (their  greatest  common  divisor),  gives  42,  the  least  common 
denominator,  and  the  required  fractions  are 

9  10 


9.  Reduce  the  fractions  —  and  —  to  others  which  have  the 
10          15 

least  common  denominator. 

D 


50  FRACTIONS. 

10.  Reduce  ^-  and  ^T-J  to  equivalent  fractions  having  the 


4ac       ,    d 
Ans.  TTT-T  and 


least  common  denominator. 


11.  Reduce  — — r  and    a    ,a  to  equivalent  fractions  having 

the  least  common  denominator. 

(a+b)*      .  c+d 
Ans.  __and__ 


PROBLEM  V. 

(87.)  To  add  fractional  quantities  together. 

RULE. 

Reduce  the  fractions  to  a  common  denominator  ;  add  the  nu- 
merators together,  and  place  their  sum  over  the  common  denom- 
inator. 

The  fractions  must  first  be  reduced  to  a  common  denomina- 
tor to  render  them  like  parts  of  unity.  Before  this  reduction, 
they  must  be  considered  as  unlike  quantities. 

EXAMPLES. 

1.  What  is  the  sum  of  -  and  -? 

*£         »> 

Reducing  to  a  common  denominator,  the  fractions  become 

3x         2x 

6  /}  * 

o 

5x 

Adding  the  numerators,  we  obtain  — . 

It  is  plain  that  three  sixths  of  x  and  two  sixths  of  x  make 
five  sixths  of  x. 

ft      f*  f> 

2.  Required  the  sum  of  ^,  ^,  and  > 

adf+bcf+bde 

bdf       '• 


FRACTIONS.  51 

3.  Required  the  sum  of  — —r  and r. 

a+b         a—b 

2a        .  a+2x 

4.  Required  the  sum  of  5x,  ^-5,  and  — - — . 

OX  437 

5.  Required  the  sum  of  2a9  3a+—,  and  #+-g-» 


6.  Required 

7.  Required 

8.  Required 
9.  Required 

Ans. 

58o: 
45  ' 

x* 

the  sum  of  •       •  and         . 

Ans.  a. 

a  a—  2m           a+2m 

the  sum  ot    ,       .     ,  and  •        -. 

ma—b         na+b 
the  sum  of  —  ;  —  and  —  ;  —  . 
m+n          m-\-n 

PROBLEM  VI. 

(88.)  To  subtract  one  fractional  quantity  from  another. 

RULE. 

Reduce  the  fractions  to  a  common  denominator,  subtract  one 
numerator  from  the  other,  and  place  their  difference  over  the 
common  denominator. 

EXAMPLES. 

23;  3x 

1.  From  —  subtract  — . 

o  o 

Reducing  to  a  common  denominator,  the  fractions  become 
103;       ,  9a; 


103J    9x     x 

Hence  TTT — T^=T^; 

15      15     15 

and  it  is  plain  that  ten  fifteenths  of  x,  diminished  by  nine  fit- 
teenths  of  x,  equals  one  fifteenth  of  x. 


52  FRACTIONS. 

I2x  3x 

2.  From  -=-  subtract  —  -. 

7  & 

9x—  4y  5x—3y 

3.  From  —  —  —  subtract  --  ^-A 

7  o 

It  must  be  remembered,  that  the  minus  sign  before  the  di- 
viding line  of  a  fraction  affects  the  quotient  (Art.  82)  ;  and 
since  a  quantity  is  subtracted  by  changing  its  sign,  the  result 
of  the  subtraction  in  this  case  is 


which  fractions  may  be  reduced  to  a  common  denominator, 
and  the  like  terms  united,  as  in  addition. 

4.  From  ^-  subtract  ^-. 
b—c  b+c 


2acx 
Ans'      " 


5.  From  2x-] — ——  subtract  x— 


ft  91 

355Z-6 


Ans. 


168 


_  x      .  x—a 

6.  From  3x+^-i  subtract  x . 

2b  c 

a+b  a—b 

7.  From  — —  subtract  — ^— . 

&  iii 

13<z-5&  7a-26 

8.  From . subtract  — - — . 

A  t) 

25<z-116 


12 


Ans. 

PROBLEM  VII. 

(89.)  To  multiply  fractional  quantities  together. 

RULE. 

Multiply  all  the  numerators  together  for  a  new  numerator^ 
and  all  the  denominators  together  for  a  new  denominator. 

Let  it  be  required  to  multiply  r  by  -^ 


FRACTIONS.  53 

First,  let  us  multiply  j-  by  c.  According  to  Remark  first  of 
Art.  81,  the  product  must  be  -=-. 

But  the  proposed  multiplier  was  -\;  that  is,  we  have  used  a 
multiplier  d  times  too  great.  We  must  therefore  divide  the 
result  -j-  by  d;  and,  according  to  Remark  second  of  Art.  81, 

we  obtain 

ac 
bd' 

which  result  conforms  to  the  Rule  above  given. 

EXAMPLES. 
X          %X 

1.  Multiply  -  by  ~. 


A 

Ans.  -. 


i*    4-7*  1 0*7* 

2.  What  is  the  continued  product  of*-,  — ,  and ? 

2    5  21 

3.  Multiply  -  by  — . 

*  J  a    J  a+c 

4.  What  is  the  continued  product  of  — , ,  and  — r-  ? 

CL        C  <iiU 


Ans.  Sax. 


5.  Multiply  b+^  by  -. 


6.  Multiply  ^-  by  £±1. 

be       J    b+c 

Ans.  T^ 

7.  What  is  the  continued  product  of  x,- — ,  and  ^—7 


a 

- 

Ans. 


54  FRACTIONS. 


Ans. 

(90.)  Ex.  I.  Multiply  —  by  -. 

According  to  the  preceding  Article,  the  result  must  be  -^. 
But,  according  to  Art.  68,  —  may  be  written  or* ;  -5  may  be 

written  a"3 ;  and  — j  may  be  written  or6. 

Therefore,  a-^Xa-*=ar*. 

That  is,  the  Rule  of  Art.  50  is  general,  and  applies  to  nega- 
tive as  well  as  positive  exponents. 
Ex.  2.  Multiply  -fr-a  by  b~*. 

Ans.  -&-* 

3.  Multiply  a-3  by  a9. 

4.  Multiply  ftr-»  by  b*. 

5.  Multiply  or"  by  or". 

6.  Multiply  6-"  by  bm. 

7.  Multiply  (a-&)5  by  (a-b)-3. 


PROBLEM  VIII. 
(91.)  To  divide  one  fractional  quantity  by  another. 

RULE. 

Invert  the  divisor,  and  proceed  as  in  multiplication. 

If  the  two  fractions  have  the  same  denominator,  then  the 
quotient  of  the  fractions  will  be  the  same  as  the  quotient  of 
their  numerators. 

3  9 

Thus  it  is  plain  that  —  is  contained  in  —  as  often  as  3  is 

contained  in  9. 

But  when  the  two  fractions  have  not  the  same  denominator, 
we  must  reduce  them  to  this  form  by  Problem  IV. 

ft  f*  '     ' 

Let  it  be  required  to  divide  T  by  ^. 


FRACTIONS.  55 

Reducing  to  a  common  denominator,  we  have  j-j  to  be  di- 

,  ,      be 
videdby^. 

It  is  now  plain  that  the  quotient  must  be  represented  by  the 
division  of  ad  by  be,  which  gives 

ad 
be 

the  same  result  as  obtained  by  the  above  Rule. 
a    c     a    d    ad 

ThuS,  —-nr-X-^T-. 

b     d    b     c     be 

EXAMPLES. 

1.  Divide  |  by  -£. 

Ans.  1$. 

2a .     4c 

2.  Divide  -r  by  -=-. 

b         d 

2#a  x 

3.  Divide  -5—5  by 


a*+x*    J  x+a 

x+1  .     2x 

4.  Divide  by  -^-. 

t>  3 

,  x— b        3cx 

5.  Divide  -r— -  by  -7-7. 

Scd     J   4d 


Ans. 

^b>~- 

^^Jj  G^^.6 


0    TV  -i 
6.  Btvide 

c 


'    Tk-   -j       a          b  a          b 

7.  Divide  — -H r  by r -7. 

a+b    a—b    J  a—b    a+b 

Ans.  Unity. 

(92.)  Ex.  1.  Divide  i  by  —t. 
According  to  the  Rule  of  the  preceding  Article,  we  have 


56  FRACTIONS. 

But  ~i  may  be  written  or6  ;  —  may  be  written  a—3  ;  and  — 

is  equal  to  or3. 

Hence  a-5+a-^=a-\ 

That  is,  the  Rule  of  Art.  66  is  general,  and  applies  to  nega- 
tive as  well  as  positive  exponents. 

Ex.  2.  Divide  —  ZH5  by  —  b~\ 

Ans.  ft-3. 

3.  Divide  a8  by  or*. 

4.  Divide  1  by  a—4. 

5.  Divide  6an  by  —  Sa-3. 

6.  Divide  #"-*  by  bm. 

7.  Divide  I2x~  3y—4  by  —4xy*. 

8.  Divide  (x—  y)~*  by  (x—  y)-6. 

(93.)  According  to  the  definition,  Art.  33,  the  reciprocal  of 
a  quantity  is  the  quotient  arising  from  dividing  a  unit  by  that 
quantity. 

Hence  the  reciprocal  of  j- 

a          b    b 

is  l-f~=lx-=-. 

b          a    a 

That  is,  the  reciprocal  of  a  fraction  is  the  fraction  inverted. 

a 


Thus  the  reciprocal  of  •?—.  —  is  -  -. 
b+x        a 

The  reciprocal  of  ^—  is  b+c. 

Hence,  to  divide  by  any  quantity  is  the  same  as  to  multiply 
by  its  reciprocal,  and  to  multiply  by  any  quantity  is  the  same 
as  to  divide  by  its  reciprocal. 

(94.)  The  numerator  or  denominator  of  a  fraction  may  be 
itself  a  fraction  ; 

.    f 

As  b  or  - 

c 

c    d 

Such  expressions  are  easily  reduced  by  applying  the  pre- 
ceding principles. 


FRACTIONS.  57 


(-} 

Thus,  \bJ  means  r-=-, 


which,  according  to  Remark  second,  Art.  81,  equals  —  . 

oc 

—  b 

Again,  /b\  means  a-:--, 

I  -  I  c 

\c/ 

<zc 
which,  according  to  Art.  91,  equals  - 


Also,  -r^r  means  the  same  as  ?_^£ 

W 
which,  according  to  Art.  91,  equals  -7-. 

3. 

^.  1.  Find  the  value  of  the  fraction  f. 

2- 

jBa;.  2.  Find  the  value  of  the  fraction  -A 


SECTION  VII. 


SIMPLE  EQUATIONS. 

(95.)  An  equation  is  a  proposition  which  declares  the  equality 
of  two  quantities  expressed  algebraically. 

Thus,  x—  4=6—  x,  is  a  proposition  expressing  the  equality 
of  the  quantities  #—4  and  b—  x. 

The  quantity  on  the  left  side  of  the  sign  of  equality  is  called 
the  first  member  of  the  equation;  the  quantity  on  the  right,  the 
second  member. 

Equations  are  usually  composed  of  certain  quantities  which 
are  known,*zud  others  which  are  unknown.  The  known  quan- 
tities are  represented  either  by  numbers  or  by  the  first  letters 
of  the  alphabet,  a,  b,  c,  &c.  ;  the  unknown  quantities  by  the  last 
letters,  x,  y,  z,  &c. 

An  identical  equation  is  one  in  which  the  two  members  are 
identical,  or  may  be  reduced  to  identity  by  performing  the  op- 
erations which  are  indicated  in  them. 

Thus,  2x-5=2x-5 


(x+y)  (x-y)=x*-y\ 

A  root  of  an  equation  is  the  value  of  the  unknown  quantity 
in  the  equation. 

(96.)  Equations  are  divided  into  degrees,  according  to  the 
highest  power  of  the  unknown  quantity  which  they  contain. 

Those  which  contain  only  the  first  power  of  the  unknown 
quantity  are  called  simple  equations,  or  equations  of  the  first 
degree. 

As  ax-\-b=cx+d. 

Those  in  which  the  highest  power  of  the  unknown  quantity 


SIMPLE    EdUATIONS.  59 

is  a  square,  are  called  quadratic  equations,  or  equations  of  the 
second  degree. 

As  4x*-2x=5-x\ 

Those  in  which  the  highest  power  is  a  cube,  are  called  cubic 
equations,  or  equations  of  the  third  degree. 

As  x*+px*=2q. 

So,  also,  we  have  biquadratic  equations,  or  equations  of  the 

fourth  degree  ;  equations  of  the  fifth,  sixth, nth 

degree. 

Thus,  xn+pxn—l=r,  is  an  equation  of  the  nth  degree. 

In  general,  the  degree  of  an  equation  is  determined  by  the 
highest  of  the  exponents  with  which  the  unknown  quantity  is  af- 
fected. 

(97.)  Numerical  equations  are  those  which  contain  only  par- 
ticular numbers,  with  the  exception  of  the  unknown  quantity, 
which  is  always  denoted  by  a  letter. 

Thus,  x*+4:X*=3x+12  is  a  numerical  equation. 

Literal  equations  are  those  in  which  the  known  quantities 
are  represented  by  letters,  or  by  letters  and  numbers. 

Thus,  x*+px*+qx=r        ) 

40     sir     2     e  f  are  literal  equations. 
x*— 3pxs+5qx*=5  ) 

To  solve  an  equation  is  to  find  the  value  of  the  unknown 
quantity,  or  to  find  a  number  which,  substituted  for  the  un- 
known quantity  in  the  equation,  renders  the  first  member 
identical  with  the  second. 

The  difficulty  of  solving  equations  depends  upon  their  de- 
gree, and  the  number  of  unknown  quantities.  We  will  begin 
with  the  most  simple  case. 

SIMPLE  EQUATIONS  CONTAINING  BUT  ONE  UNKNOWN  QUAN- 
TITY. 

(98.)  The  various  operations  which  we  perform  upon  equa- 
tions in  order  to  deduce  the  value  of  the  unknown  quantities, 
are  founded  upon  the  following  principles : 

1.  If  to  two  equal  quantities  the  same  quantity  be  added,  the 
sums  will  be  equal. 

2.  If  from  two  equal  quantities  the  same  quantity  be  sub- 
tracted, the  remainders  will  be  equal. 

3.  If  two  equal  quantities  be  multiplied  by  the  same  quan- 
tity, the  products  will  be  equal. 


60  SIMPLE    EQUATIONS. 

4.  If  two  equal  quantities  be  divided  by  the  same  quantity, 
the  quotients  will  be  equal. 

(99.)  The  unknown  quantity  may  be  combined  with  the 
known  quantities  in  the  given  equation  by  the  operations  of 
addition,  subtraction,  multiplication,  or  division. 

We  shall  consider  these  different  cases  in  succession. 

I.  The  unknown  quantity  may  be  combined  with  known 
quantities  by  addition. 

Let  it  be  required  to  solve  the  equation 


If  from  the  two  equal  quantities,  x+6  and  24,  we  subtract 
the  same  quantity  6,  the  remainders  will  be  equal,  according 
to  the  last  Article,  and  we  shall  have 

#+6-6=24-6, 
or  #=24—  6, 

=  18,  the  value  of  x  required. 
So,  also,  in  the  equation 

x+a=b, 

subtracting  a  from  each  of  the  equal  quantities,  x  +  a  and  b,  the 
result  is 

x=b—a,  the  value  of  x  required. 

(100.)  II.  The  unknown  quantity  may  be  combined  with 
known  quantities  by  subtraction. 
Let  the  equation  be 

a;_  6=24. 

If  to  the  two  equal  quantities,  x—  6  and  24,  the  same  quan- 
tity 6  be  added,  the  sums  will  be  equal,  according  to  Art.  98, 
and  we  have 

^-6+6=24+6, 

or  a;  =30,  the  value  of  a;  required. 
So,  also,  in  the  equation 

#—  a=bt 

adding  a  to  each  of  these  equal  quantities,  the  result  is 

x=b+a,  the  value  of  a;  required. 
From  the  preceding  examples,  it  follows  that 
We  may  transpose  any  term  of  an  equation  from  one  member 

to  the  other  by  changing  its  sign. 

We  may  change  the  sign  of  every  term  of  an  equation  with- 

out destroying  the  equality. 


SIMPLE    EQUATIONS.  61 

This  is,  in  fact,  the  same  thing  as  transposing  every  term  in 
each  member  of  the  equation. 

If  the  same  quantity  appear  in  each  member  of  the  equation 
affected  with  the  same  sign,  it  may  be  suppressed. 

(101.)  III.  The  unknown  quantity  may  be  combined  with 
known  quantities  by  multiplication. 

Let  the  equation  be 

6#=24. 

If  we  divide  each  of  the  equal  quantities,  6x  and  24,  by  the 
same  quantity  6,  the  quotients  will  be  equal,  and  we  shall  have 

_24 

"¥' 

=4,  the  value  of  x  required. 
So,  also,  in  the  equation 

ax=b, 
dividing  each  of  these  equals  by  a,  the  result  is 

#=-,  the  value  of  x  required. 

From  this  it  follows,  that 

When  the  unknown  quantity  is  multiplied  by  a  known  quan- 
tity, the  equation  is  solved  by  dividing  both  members  by  this 
known  quantity. 

(102.)  IV.  The  unknown  quantity  may  be  combined  with 
known  quantities  by  division. 

Let  the  equation  be 


X 

If  we  multiply  each  of  the  equal  quantities,  -  and  24,  by  the 

same  quantity  6,  the  products  will  be  equal,  and  we  shall  have 

x=l44,  the  value  of  a;  required. 
So,  also,  in  the  equation 

a 

multiplying  each  of  these  equals  by  a,  the  result  is 

x=ab,  the  value  of  x  required. 
From  this  it  follows,  that 
When  the  unknown  quantity  is  divided  by  a  known  quantity. 


<>2  SIMPLE    EQUATIONS. 

the  equation  is  solved  by  multiplying  both  members  by  this  known 
quantity. 

(103.)  V.  Several  terms  of  an  equation  may  be  fractional. 

Let  the  equation  be 

?—  ?     4 
2~3+5* 

Multiplying  each  of  these  equals  by  2,  the  result  is 

48 
*=3+5- 
Multiplying  each  of  these  last  equals  by  3,  we  obtain 


and  multiplying  again  by  5,  we  obtain 
15*=20+24, 
an  equation  free  from  fractions. 

We  might  have  obtained  the  same  result  by  multiplying  the 
original  equation  at  once  by  the  product  of  all  the  denom- 
inators. 

Thus,  multiplying  by  2X3X5,  we  have 
30z_60     120 
~2~-~3~+~5~' 
or  reducing,  we  have 

15z=20+24,  as  before. 
So,  also,  in  the  equation 

x_b    d 
a^c+ef 

multiplying  successively  by  all  the  denominators,  or  by  a  c  e 
at  once,  we  obtain 

acex_dbce    acde 
a          c         e    ' 

Canceling  from  each  term  the  letter  which  is  common  to  its 
numerator  and  denominator,  we  have 
cex=abe+acd, 

an  equation  clear  of  fractions. 
Hence  it  appears  that 

An  equation  may  be  cleared  of  fractions  by  multiplying  each 
member  into  all  the  denominators. 

(104.)  From  the  preceding  remarks,  we  deduce  the  fol- 
lowing 


SIMPLE    EQUATIONS.  (53 


RULE  FOB  THE  SOLUTION  OF  A  SIMPLE  EQUATION  CONTAINING 
ONE  UNKNOWN  QUANTITY. 

1.  Clear  the  equation  of  fractions,  and  perform  in  both  mem- 
bers all  the  algebraic  operations  indicated. 

2.  Transpose  all  the  terms  containing  the  unknown  quantity 
to  one  side,  and  all  the  remaining  terms  to  the  other  side  of  the 
equation,  and  reduce  each  member  to  its  most  simple  form. 

3.  Divide  each   member  by  the   coefficient  of  the  unknown 
quantity. 

EXAMPLES. 

1.  Given  5x+  8=4^  +  10,  to  find  the  value  of  x. 
Transposing  4x  to  the  first  member  of  the  equation,  and  8 

to  the  second  member,  taking  care  to  change  their  signs  (Art. 
100),  we  have 

5^-4^=10-8. 

Uniting  similar  terms,  x=2. 

In  order  to  verify  this  result,  put  2  in  the  place  of  x  wher- 
ever it  occurs  in  the  original  equation,  and  we  shall  obtain 

5X2+8=4X2  +  10. 
That  is,  10+8=8  +  10, 

or  18=18, 

an  identical  equation,  which  proves  that  we  have  found  the 
correct  value  of  x. 

IT       3T 

2.  Given  x—  7=  -  +  0,  to  find  the  value  of  x. 

O       o 

Multiplying  every  term  of  the  equation  by  5  and  also  by  3, 
in  order  to  clear  it  of  fractions  (Art.  103),  we  obtain 


Hence,  by  transposition, 


or  7z=105, 

and  therefore  x=—-  =  15. 

To  verify  this  result,  put  15  in  the  place  of  x  in  the  original 
equation,  and  we  have 

_15     15 

n     ~T+y 

4 


64  SIMPLE    EQUATIONS. 

That  is,  15-7=3+5, 

or  8=8, 

an  identical  equation. 

3.  Given  Sax— 4ab=2ax— 6ac,  to   find  the  value  of  x  in 
terms  of  b  and  c. 

Dividing  every  term  by  a,  we  have 

3x-4b=2x-6c. 
By  transposition, 

or  x=4b— 6c. 

This  result  may  be  verified  in  the  same  manner  as  the 
ceding. 

4.  Given  3#a—  10x=8x+x*,  to  find  the  value  of  x. 

Ans.  x=9. 

a(d*+x*)          ,  ax 

5.  Given  -^— = L=ac+-j,  to  find  x. 

dx  d 


6.  Given  — j— +6#= — ~ — ,  to  find 


7.  Given  —  =bc+d+-,  to  find  a;. 
a:  x 


.  #=-. 
c 


Ans.  x=9. 


,  ^    t*  -W7 

,  —  '  6        ,  U*-37  4    fi    , 

8.  Given  3a;H — =5H ,  to  find  x. 

Oi  i{ n*4g      /  if.  ~? 

9.  Given  5ax—2b+4bx=2x+5c,  to  find  #. 

J.WS.    .T  = 


5a+4b-2' 

Qg.  _  g  g^»  _  £ 

10.  Given  #H  --  -  —  =12  --  -  —  ,  to  find  the  value  of#. 

Ans.  .T=5. 

Sz-ll     5x—  5     97—  7x 

11.  Given  21  +—^—=—5—+  —  -  —  ,  to  find  x. 

ID  o  Z 

(105.)  An  equation  may  always  be  cleared  of  fractions  by 
multiplying  each  member  into  all  the  denominators  according 


SIMPLE    EaUATIONS.  65 

10  Art.  103.     But  sometimes  the  same  object  may  be  attained 
by  a  less  amount  of  multiplication. 

Thus,  in  the  preceding  example,  the  equation  maybe  cleared 
of  fractions  by  multiplying  each  term  by  16,  instead  of  16X8 
X2,  and  it  is  important  to  avoid  all  useless  multiplication.  In 
general,  it  is  sufficient  to  multiply  by  the  least  common  multiple 
of  all  the  denominators.  See  Art.  86. 

12.  Given  3z-^- 4=— y^,  to  find  x. 

13.  Given  3x— a+cx=— — ,  to  find  x. 

o  (I 


Ans.  x— 


8a+3ac-& 


Oj.  y>  Oj* 

14.  Given c-f  r=4oH — T,  to  find  x. 

a  b  a 


abed 
Ans.  x=- 


3bd+ ad—  4abd— 2ab' 

afc 

15.  Given  (a+x)  (b+x)-a(b+c)=-^-+x\  to  find  x. 

ac 

Ans.  x=—. 
o 

ll-3x     4x+2  7«+14 

16.  Given — =5— 6oH — ,  to  find  x. 

DO  O 

3x-3          20-x     Gx-S     4x-4 

17.  Given  x — \-4= — 1 — ,  to  find  x. 

o  fi  7  o 

7a;+16      x+S      x 

18.  dven  -^ 5— fTg.  ^  find  x. 

ftc+7     7Z-13     2x+4 

19.  G,ven        -+--=  -jj-,  to  find  x. 


20.  Given  -ab+-ac— -cx=-ac-{-2ab— 6cx,  to  find  the  value 
o         5          o         4 

of  x. 

Ans.  x= 


320c 


SOLUTION  OF  PROBLEMS. 

(106.)  The  solution  of  a  Problem  by  Algebra  consists  of 
two  distinct  parts : 


66  SIMPLE    EQUATIONS. 

1.  To  express  the  conditions  of  the  problem  algebraically ; 
that  is,  to  form  the  equation. 

2.  To  solve  the  equation. 

The  second  operation  has  already  been  explained,  but  the 
first  is  often  more  embarrassing  to  beginners  than  the  second. 
Sometimes  the  statement  of  a  problem  furnishes  the  equation 
directly;  and  sometimes  it  is  necessary  to  deduce  from  the 
statement  new  conditions,  which  are  to  be  expressed  alge- 
braically. The  former  are  called  explicit  conditions  ;  and  those 
which  are  deduced  from  them,  implicit  conditions. 

It  is  impossible  to  give  a  general  rule  which  will  enable  us 
to  translate  every  problem  into  algebraic  language.  The 
power  of  doing  this  with  facility  can  only  be  acquired  by  re- 
flection and  practice. 

The  following  directions  may  be  found  of  some  service. 

Denote  one  of  the  required  quantities  by  x  ;  then,  by  means 
of  this  letter,  with  the  algebraic  signs,  perform  the  same  opera- 
tions which  would  be  necessary  to  verify  its  value  if  it  was  al- 
ready known. 

Problem  1.  What  number  is  that,  to  the  double  of  which  if 
16  be  added,  the  sum  is  equal  to  four  times  the  required  num- 
ber ? 

Let  x  represent  the  number  required. 

The  double  of  this  will  be  2x. 

This  increased  by  16  should  equal  4x. 

Hence,  by  the  conditions,  2x-\-lQ=4x. 

The  problem  is  now  translated  into  algebraic  language,  and 
it  only  remains  to  solve  the  equation  in  the  usual  way. 

Transposing,  we  obtain 

I6=4x— 2x=2x, 

and  8= x, 

or  x=S. 

To  verify  this  number,  we  have  but  to  double  8,  and  add 
16  to  the  result ;  the  sum  is  32,  which  is  equal  to  four  times  8, 
according  to  the  conditions  of  the  problem. 

Prob.  2.  What  number  is  that,  the  double  of  which  exceeds 
its  half  by  6? 

Let  x  =  the  number  required. 

Then,  by  the  conditions, 


SIMPLE    EdUATIONS.  67 


Clearing  of  fractions, 

4x-x=I2, 
or  3x=12. 

Hence  x=4. 

To  verify  this  result,  double  4,  which  makes  8,  and  dimmish 
,t  by  the  half  of  4,  or  2  ;  the  result  is  6,  according  to  the  con- 
ditions of  the  problem. 

Prob.  3.  The  sum  of  two  numbers  is  8,  and  their  difference 
2.  What  are  those  numbers  ? 

Let  x  =  the  least  number. 

Then  x+2  will  be  the  greater  number. 

The  sum  of  these  is  2z+2,  which  is  required  to  equal  8. 
Hence  we  have 

2x+2=8. 

By  transposition,  2x=S—  2=6, 

and  x=3,  the  least  number. 

Also,  x+2  =5,  the  greater  number. 

Verification.  5+3=8  ) 

5_o_o  (  according  to  the  conditions. 

The  following  is  a  generalization  of  the  preceding  Problem. 

Prob.  4.  The  sum  of  two  numbers  is  «,  and  their  difference 
b.  What  are  those  numbers  ? 

Let  x  represent  the  least  number. 

Then  x  +b  will  represent  the  greater  number. 

The  sum  of  these  is  2x+b,  which  is  required  to  equal  a. 

Hence  we  have 

2x+b=a. 
By  transposition,  2x=a—b, 

a  —  b     a    b 
or  x=  ~~7r~=o~o»  tne  *ess  numker. 

a     b          a     b 
Hence  x+b=-—~+b=  x+-,  the  greater  number. 

/«      <*  £      Z 

As  these  results  are  independent  of  any  particular  value  at- 
tributed to  the  letters  a  and  b,  it.  follows  that 

Half  the  difference  of  two  quantities,  added  to  half  their  sum, 
is  equal  to  the  greater  ;  and 


68  SIMPLE    EQ,UATIONS. 

Half  the  difference  subtracted  from  half  the  sum  is  equal  to 
the  less. 

The  expressions  -+-  and  -— -  are  called  formulas,  because 

&      A  £      £ 

they  may  be  regarded  as  comprehending  the  solution  of  all 
questions  of  the  same  kind;  that  is,  of  all  problems  in  which 
we  have  given  the  sum  and  difference  of  two  quantities. 

Thus,  let  a=S  j  ag  ^  the  preceding  problem. 
o — <ii  ) 

Then  -+-=—-— =5,  the  greater  number. 

22  A 

. a     b     8—2 

And  -—-=—— =3,  the  less  number. 
22        <« 

^     =10;  their  difference  =  6;  required  the  numbeis. 
|J  12  "  2  " 

23  "  11 

•5  §          100  «  50  " 

§  g          100  "  1 

K.      j> 

•S  **    .  R  «  I  « 

O<~  5  2 

10  "  i  " 

Prob.  5.  From  two  towns  which  are  54  miles  distant,  two 
travelers  set  out  at  the  same  time  with  an  intention  of  meet- 
ing. One  of  them  goes  4  miles  and  the  other  5  miles  per  hour. 
In  how  many  hours  will  they  meet  ? 

Let  x  represent  the  required  number  of  hours. 

Then  4x  will  represent  the  number  of  miles  one  traveled, 

and  5x  the  number  the  other  traveled ; 

and  since  they  meet,  they  must  together  have  traveled  the 
whole  distance. 

Consequently,  4#+5#=54. 

Hence  9#=54, 

or  x=6. 

Proof.  In  6  hours,  at  4  miles  an  hour,  one  would  travel  24 
miles ;  the  other,  at  5  miles  an  hour,  would  travel  30  miles. 
The  sum  of  24  and  30  is  54  miles,  which  is  the  whole  distance. 

This  Problem  may  be  generalized  as  follows : 

Prob.  6.  From  two  points  which  are  a  miles  apart,  two 
bodies  move  toward  each  other,  the  one  at  the  rate  of  m  miles 


SIMPLE    EQUATIONS.  09 

per  hour,  the  other  at  the  rate  of  n  miles  per  hour.     In  how 
many  hours  will  they  meet? 

Let  x  represent  the  required  number  of  hours. 

Then  mx  will  represent  the  number   of  miles  one  body 
moves, 

and  nx  the  miles  the  other  body  moves, 
and  we  shall  obviously  have 


Hence  x=  -  . 
m-\-n 

This  is  a  general  formula,  comprehending  the  solution  of  all 
problems  of  this  kind.  Thus, 

c  0      =150;  ^  6;  £  4  ^ 

^g  90  ||  8  -I  7  Si 

-£>    >  O  —  •  J3 

135  0  |  15  9  12         'g  n 

210  20  -5  15 

Required  the  time  of  meeting. 

We  see  that  an  infinite  number  of  problems  may  be  pro- 
posed, all  similar  to  Prob.  5  ;  but  they  are  all  solved  by  the 
formula  of  Prob.  6.  We  also  see  what  is  necessary  in  order 
that  the  answers  may  be  obtained  in  whole  numbers.  The 
given  distance  (a)  must  be  exactly  divisible  by  m+n. 

Prob.  7.  A  gentleman  meeting  three  poor  persons,  divided 
60  cents  among  them  ;  to  the  second  he  gave  twice,  and  to 
the  third  three  times  as  much  as  to  the  first.  What  did  he 
give  to  each? 

Let  x  =  the  sum  given  to  the  first. 

Then  2x  =  the  sum  given  to  the  second, 
and  3x  =  the  sum  given  to  the  third. 

Then,  by  the  conditions, 


That  is,  6z=60, 

or  #=10. 

Therefore  he  gave  10,  20,  and  30  cents  to  them  respectively. 
The  learner  should  verify  this,  and  all  the  subsequent  results. 

The  same  problem  generalized. 

Prob.  8.  Divide  the  number  a  into  three  such  parts,  that  the 


70  SIMPLE    EQUATIONS. 

second  may  be  m  times,  and  the  third  n  times  as  great  as  the 
first. 

a  ma  na 

1+m+n'  1+m+n'  l+m+n 

What  is  necessary  in  order  that  the  preceding  values  may 
be  expressed nn  whole  numbers? 

Prob.  9.  A  bookseller  sold  10  books  at  a  certain  price,  and 
afterward  15  more  at  the  same  rate.  Now  at  the  last  sale  he 
received  25  dollars  more  than  at  the  first.  What  did  he  re- 
ceive for  each  book  ? 

Ans.  Five  dollars. 
The  same  Problem  generalized. 

Prob.  10.  Find  a  number  such  that  when  multiplied  success- 
ively by  m  and  by  n,  the  difference  of  the  products  shall  be  a. 

Ans.  —  — . 
m— n 

Prob.  11.  A  gentleman  dying,  bequeathed  1000  dollars  to 
three  servants.  A  was  to  have  twice  as  much  as  B,  and  B 
three  times  as  much  as  C.  What  were  their  respective 
shares  ? 

Ans.  A  received  $600,  B  $300,  and  C  $100. 
Prob.  12.  Divide  the  number  a  into  three  such  parts  that  the 
second  may  be  m  times  as  great  as  the  first,  and  the  third  n 
times  as  great  as  the  second. 

a  .  ma  mna 

A  ft  O -    •      •  -    _._— r 

l+m+mn9  l+m+mnj   l+m+mn' 

Prob.  13.  A  hogshead  which  held  120  gallons  was  filled 
with  a  mixture  of  brandy,  wine,  and  water.  There  were  10 
gallons  of  wine  more  than  there  were  of  brandy,  and  as  much 
water  as  both  wine  and  brandy.  What  quantity  was  there  of 
each  ? 

Ans.  Brandy  25  gallons,  wine  35,  and  water  60  gallons. 
Prob.  14.  Divide  the  number  a  into  three  such  parts,  that 
the  second  shall  exceed  the  first  by  m,  and  the  third  shall  be 
equal  to  the  sum  of  the  first  and  second. 

a— 2m     a+2m     a 

ns'  ~T~'  ~T~;  2* 

Prob.  15.  A  person  employed  four  workmen,  to  the  first  of 
whom  he  gave  2  shillings  more  than  to  the  second  ;  to  the 


SIMPLE    EQUATIONS.  71 

second  3  shillings  more  than  to  the  third  ;  and  to  the  third  4 
shillings  more  than  to  the  fourth.  Their  wages  amount  to  32 
shillings.  What  did  each  receive  ? 

Ans.  They  received  12,  10,  7,  and  3  shillings  respectively. 

Prob.  16.  Divide  the  number  a  into  four  such  parts,  that  the 
second  shall  exceed  the  first  by  m,  the  third  shall  exceed  the 
second  by  n,  and  the  fourth  shall  exceed  the  third  by  p. 

a—3m—2n—p     a+m—2n—p 
Ans.  .      __     P.  .      __L, 

a+m+2n—p      a+m+2n+3p 
~4~  ~~4~ 

(107.)  Problems  which  involve  several  unknown  quantities 
may  often  be  solved  by  the  use  of  a  single  unknown  letter. 
Most  of  the  preceding  examples  are  of  this  kind.  In  general, 
when  we  have  given  the  sum  or  difference  of  two  quantities, 
both  of  them  may  be  expressed  by  means  of  the  same  letter. 
For  the  difference  of  two  quantities  added  to  the  less  must  be 
equal  to  the  greater  ;  and  if  one  of  two  quantities  be  sub- 
tracted from  their  sum,  the  remainder  will  be  equal  to  the 
other. 

Prob.  17.  At  a  certain  election  36000  votes  were  polled  ; 
and  the  candidate  chosen  wanted  but  3000  of  having  twice  as 
many  votes  as  his  opponent.  How  many  voted  for  each  ? 

Let  x  =  the  number  of  votes  for  the  unsuccessful  candidate 

Then  36000—  x  =  the  number  the  successful  one  had, 

And  36000-o;+3000=2;r. 

Ans.  13000  and  23000. 

Prob.  18.  Divide  the  number  a  into  two  such  parts,  that  one 
part  increased  by  b  shall  be  equal  to  m  times  the  other  part. 

ma—b     a-\-b 


m+l 

Prob.  19.  A  train  of  cars  moving  at  the  rate  of  20  miles  per 
hour,  had  been  gone  three  hours,  when  a  second  train  followed 
at  the  rate  of  25  miles  per  hour.  In  what  time  will  the  second 
train  overtake  the  first  ? 

Let  x  =  the  number  of  hours  the  second  train  is  in  motion, 

x-}-3  =  the  time  of  the  first  train. 

Then  25x  —  the  number  of  miles  traveled  by  the  second  train, 
2Q(x-\-3)  =  the  miles  traveled  by  the  first  train. 


72  SIMPLE    EQUATIONS. 

But  at  the  time  of  meeting  they  must  both  have  traveled  the 
same  distance. 

Therefore  25z=20z+60. 

By  transposition,  5#=60, 
and  x=l2. 

Proof.  In  12  hours,  at  25  miles  per  hour,  the  second  train 
goes  300  miles  ;  and  in  15  hours,  at  20  miles  per  hour,  the  first 
train  also  goes  300  miles  ;  that  is,  it  is  overtaken  by  the  sec- 
ond train. 

Prob.  20.  Two  bodies  move  in  the  same  direction  from  two 
places  at  a  distance  of  a  miles  apart  ;  the  one  at  the  rate  of  n 
miles  per  hour,  the  other  pursuing  at  the  rate  of  m  miles  per 
hour.  When  will  they  meet  ? 

Ans.  In  -  hours. 
w—  n 

This  Problem,  it  will  be  seen,  is  essentially  the  same  as 
Prob.  10. 

Prob.  21.  Divide  the  number  197  into  two  such  parts,  that 
four  times  the  greater  may  exceed  five  times  the  less  by  50. 

Ans.  82  and  115. 

Prob.  22.  Divide  the  number  a  into  two  such  parts,  that  m 
times  the  greater  may  exceed  n  times  the  less  by  b. 

ma—b     na+b 

Ans- 


When  7i=l,  this  Problem  reduces  to  Problem  18. 
When  b=Q,  this  Problem  reduces  to  Problem  24. 

Prob.  23.  A  prize  of  2329  dollars  was  divided  between  two 
persons,  A  and  B,  whose  shares  were  in  the  ratio  of  5  to  12. 
What  was  the  share  of  each  ? 

Beginners  almost  invariably  put  x  to  represent  one  of  the 
quantities  sought  in  a  problem  ;  but  a  solution  may  often  be 
very  much  simplified  by  pursuing  a  different  method.  Thus, 
in  the  preceding  problem,  we  may  put  x  to  represent  one  fifth 
of  A's  share.  Then  5x  will  be  A's  share,  and  I2x  will  be  B's, 
and  we  shall  have  the  equation 


and  x  =137, 

consequently  their  shares  were  685  and  1644  dollars. 


SIMPLE    EQUATIONS. 

Prob.  24.  Divide  the  number  a  into  two  such  parts,  that  the 
first  part  may  be  to  the  second  as  m  to  n. 

ma         na 
Ans.  — ; —  ;  — • — . 
m+n    m-\-n 

Prob.  25.  What  number  is  that  whose  third  part  exceeds  its 
fourth  part  by  16? 

Let  I2x  =  the  number. 
Then  4z-3x=16, 

or  x=lQ. 

Therefore  the  number  =  12X  16=192. 

Prob.  26.  Find  a  number  such  that  when  it  is  divided  suc- 
cessively by  m  and  by  n,  the  difference  of  the  quotients  shali 
be  a. 

.         mna 

Ans.  . 

n— m 

Prob.  27.  What  two  numbers  are  as  2  to  3,  to  each  of 
which,  if  four  be  added,  the  sums  will  be  as  5  to  7  ? 

A  strict  adherence  to  system  would  have  required  this  ex- 
ample to  be  placed  after  the  subject  of  Proportion,  which  is 
treated  of  in  Section  XIII.  It  is,  however,  only  necessary  to 
assume  one  simple  principle  which  is  employed  in  Arithmetic, 
viz.,  If  four  quantities  are  proportional,  the  product  of  the  ex- 
tremes is  equal  to  the  product  of  the  means. 

Thus,  if  a:b::c:d. 

Then  ad=bc. 

In  the  preceding  Problem,  let  2x  and  3x  be  the  numbers. 
Then  2x+4  :  Sx+4  : :  5  :  7, 

and  by  the  last  principle, 

1 4^+28  =  15«+20. 

Prob.  28.  What  two  numbers  are  as  m  to  n,  to  each  of 
which,  if  a  be  added,  the  sums  shall  be  as  p  to  q  ? 

ma(p-q)  _  na(p-q) 

J\.'HS.  •  • 

mq—np         mq—np 

Prob.  29.  A  gentleman  divides  a  dollar  among  12  children, 
giving  to  some  9  cents  each,  and  to  the  rest  7  cents.  Ho\v 
many  were  there  of  each  class  ? 

Prob.  30.  Divide  the  number  a  into  two  such  parts,  that  if 


74  SIMPLE    EaUATIONS. 

the  first  is  multiplied  by  m  and  the  second  by  n,  the  sum  of 
the  products  shall  be  b. 

b  —  na     ma  —  b 
Ans.  -  ;  -  . 
m—  n      m—n 

Prob.  31.  If  the  sun  moves  every  day  one  degree,  and  the 
moon  thirteen,  and  the  sun  is  now  60  degrees  in  advance  of 
the  moon,  when  will  they  be  in  conjunction  for  the  first  time, 
second  time,  and  so  on  ? 

Prob.  32.  If  two  bodies  move  in  the  same  direction  upon  the 
circumference  of  a  circle  which  measures  a  miles,  the  one  at 
the  rate  of  n  miles  per  day,  the  other  pursuing  at  the  rate  of  m 
miles  per  day,  when  will  they  meet  for  the  first  time,  second 
time,  &c.,  supposing  them  to  be  b  miles  apart  at  starting  ? 

b         a+b     2a+b 

Ans.  In  --  ;  -  ;  --  ,  &c.,  days. 
m—n     m—n     m—n 

It  will  be  seen  that  this  Problem  includes  Prob.  20. 

Prob.  33.  Divide  the  number  12  into  two  such  parts,  that  the 
difference  of  their  squares  may  be  48. 

Prob.  34.  Divide  the  number  a  into  two  such  parts,  that  the 
difference  of  their  squares  may  be  b. 


2a          2a' 

Prob.  35.  The  estate  of  a  bankrupt,  valued  at  21000  dollars, 
is  to  be  divided  among  three  creditors  according  to  their  re- 
spective claims.  The  debts  due  to  A  and  B  are  as  2  to  3, 
while  B's  claims  and  C's  are  in  the  ratio  of  4  to  5.  What  sum 
must  each  receive  ? 

Prob.  36.  Divide  the  number  a  into  three  parts,  which  shall 
be  to  each  other  as  m  :  n  :  p. 

ma  na  pa 

\  M   O  .______—,      •          -  •         _  **      - 

m-\-n-\-p  '  m+n+p  '  m-\-n+p' 

When  p=l9  Prob.  36  leduces  to  the  same  form  as  Prob.  8. 

Prob.  37.   A  grocer  has  two  kinds  of  tea,  one  worth  72 

cents  per  pound,  the  other  40  cents.     How  many  pounds  of 

each  must  be  taken  to  form  a  chest  of  80  pounds,  which  shall 

be  worth  60  cents  ? 

Ans.  50  pounds  at  72  cents,  and  30  pounds  at  40  cents. 
Prob.  38.  A  grocer  has  two  kinds  of  tea,  one  worth  a  cents 
per  pound,  the  other  b  cents.     How  many  pounds  of  each  must 


SIMPLE    EaUATIUNS.  75 


be  taken  to  form  a  mixture  of  n  pounds,  which  shall  be  worth 
c  cents  ? 

n(c-b) 

Ans.  -  j  —  pounds  at  a  cents, 
a—  b 

,  n(a—c) 

and  —  -  —  T—  pounds  at  b  cents. 
a—  b     r 

Prob.  39.  A  can  perform  a  piece  of  work  in  6  days  ;  B  can 
perform  the  same  work  in  8  days  ;  and  C  can  perform  the 
same  work  in  24  days.  In  what  time  will  they  finish  it  if  all 
work  together? 

Prob.  40.  A  can  perform  a  piece  of  work  in  a  days,  B  in  b 
days,  and  C  in  c  days.  In  what  time  will  they  perform  it  if  all 
work  together  ? 

abc 

Ans.  -=—  --  —  r-  days. 
ab+ac+bc 

Prob.  41.  There  are  three  workmen,  A,  B,  and  C.  A  and 
B  together  can  perform  a  piece  of  work  in  27  days  ;  A  and  C 
together  in  36  days  ;  and  B  and  C  together  in  54  days.  In 
what  time  could  they  finish  it  if  all  worked  together  ? 

A  and  B  together  can  perform  ^T  of  the  work  in  one  day. 

A  and  C  "  -'_  «  one    " 

B  and  C  "  _'_  «  one    " 

Therefore,  adding  these  three  results, 

2A+2B+2C  can  perform  ^T  +  -»-+^_  in  one  day. 
=  ^•5-  in  one  day. 

Therefore,  A,  B,  and  C  together  can  perform  ^  of  the  work 
in  one  day  ;  that  is,  they  can  finish  it  in  24  days.  If  we  put 
x  to  represent  the  time  in  which  they  would  all  finish  it,  then 
they  would  together  perform  ~  part  of  the  work  in  one  day, 
and  we  should  have 

*V+*V+ik=l. 

Prob.  42.  A  and  B  can  perform  a  piece  of  labor  in  a  days  ; 
A  and  C  together  in  b  days  ;  and  B  and  C  together  in  c  days. 
In  what  time  could  they  finish  it  if  all  work  together  ? 

2abc 

Ans.  —  r-  --  —r-  days. 
ab+ac+bc      J 

This  result,  it  will  be  seen,  is  of  the  same  form  as  that  of 
Problem  40. 


76  SIMPLE    EQUATIONS. 

Prob.  43.  A  broker  has  two  kinds  of  change.  It  takes  20 
pieces  of  the  first  to  make  a  dollar,  and  4  pieces  of  the  second 
to  make  the  same.  Now  a  person  wishes  to  have  8  pieces 
for  a  dollar.  How  many  of  each  kind  must  the  broker  give 
him? 

Prob.  44.  A  has  two  kinds  of  change;  there  must  be  a 
pieces  of  the  first  to  make  a  dollar,  and  b  pieces  of  the  second 
to  make  the  same.  Now  B  wishes  to  have  c  pieces  for  a  dol- 
lar. How  many  pieces  of  each  kind  must  A  give  him  ? 

Ans.  a(c~~V  of  the  firgt  kind  .  b(a-c)  second. 

a—b  a—b 

Prob.  45.  Divide  the  number  45  into  four  such  parts,  that 
the  first  increased  by  2,  the  second  diminished  by  2,  the  third 
multiplied  by  2,  and  the  fourth  divided  by  2,  shall  all  be 
equal. 

In  solving  examples  of  this  kind,  several  unknown  quantities 
are  usually  introduced,  but  this  practice  is  worse  than  super- 
fluous.    The  four  parts  into  which  45  is  to  be  divided,  may  be 
represented  thus  : 
The  first  =x—  2, 

second  =x+2, 

third  =|, 

fourth  —2x  ; 

for  if  the  first  expression  be  increased  by  2,  the  second  dimin- 
ished by  2,  the  third  multiplied  by  2,  and  the  fourth  divided  by 
2,  the  result  in  each  case  will  be  x.  The  sum  of  the  four  parts 
is  4fc  which  must  equal  45. 

Hence  x=W. 

Therefore  the  parts  are  8,  12,  5,  and  20. 

Prob.  46.  Divide  the  number  a  into  four  such  parts,  that 
the  first  increased  by  m,  the  second  diminished  by  m,  the  third 
multiplied  by  m,  and  the  fourth  divided  by  m,  shall  all  be 
equal. 

ma  ma  a  m~a 

'  (m+l)a~     '*  (m+iy+m;  (m+iy'   (m+1)2' 


Prob.  47.  A  merchant  maintained  himself  for  three  years  at 
an  expense  of  $500  a  year;  and  each  year  augmented  that 
part  of  his  stock  which  was  not  thus  expended  by  one  third 


SIMPLE    EQUATIONS.  77 

thereof.     At  the  end  of  the  third  year  his  original  stock  was 
doubled.     What  was  that  stock  ? 

Prob.  48.  A  merchant  supported  himself  for  three  years  at 
an  expense  of  a  dollars  per  year ;  and  each  year  augmented 
that  part  of  his  stock  which  was  not  thus  expended  by  one 
third  thereof.  At  the  end  of  the  third  year  his  original  stock 
was  doubled.  What  was  that  stock  ? 

148<z 
Ans.  — . 

Prob.  49.  A  father,  aged  54  years,  has  a  son  aged  9  years. 
In  how  many  years  will  the  age  of  the  father  be  four  times 
that  of  the  son  ? 

Prob.  50.  The  age  of  a  father  is  represented  by  a,  the  age 
of  his  son  by  b.  In  how  many  years  will  the  age  of  the  fa- 
ther be  7i  times  that  of  the  son  ? 

a— nb 
Ans.  r-. 

71—1 


SECTION  VIII. 


SIMPLE  EQUATIONS   CONTAINING  TWO 
OR  MORE  UNKNOWN  QUANTITIES. 

(108.)  In  the  examples  which  have  been  hitherto  given,  each 
problem  has  contained  but  one  unknown  quantity;  or,  if  there 
have  been  more,  they  have  been  so  related  to  each  other  that 
all  have  been  expressed  by  means  of  the  same  letter.  This, 
however,  can  not  always  be  done,  and  we  are  now  to  consider 
how  equations  of  this  kind  are  resolved. 

If  we  have  two  equations,  with  two  unknown  quantities,  we 
must  endeavor  to  deduce  from  them  a  single  equation,  con- 
taining only  one  unknown  quantity.  We  must,  therefore,  make 
one  of  the  unknown  quantities  disappear,  or,  as  it  is  termed, 
we  must  eliminate  it.  There  are  three  different  methods  of 
elimination  which  may  be  practiced. 

The  first  is  by  substitution, 
"  second  "  comparison, 
"  third  "  addition  and  subtraction. 


ELIMINATION  BY  SUBSTITUTION. 

(109.)  Let  it  be  proposed  to  solve  the  system  of  equations 


x-y=  6. 

From  the  second  equation,  we  find  the  value  of  x  in  terms 
of  y,  which  gives 


Substituting  the  expression  y+6  for  x  in  the  first  equation, 
it  becomes 


SIMPLE  EQUATIONS,  ETC.  79 

from  which  we  find  that  y=3 ;  and  since  we  have  already 
seen  that  x=y+6,  we  find  that  re— 3+6=9. 

To  verify  these  values,  substitute  them  for  x  and  y  in  the 
original  equations,  and  we  shall  obtain 

9+3=12 
9-3=   6. 
Again,  take  the  equations 


5x+4y=22. 
From  the  first  equation  we  find 

I3-2x 


Substituting  this  value  of  y  in  the  second  equation,  it  becomes 

13—  2x 
5z+4X  -  -  -  =22, 

o 

an  equation  containing  only  x,  which,  when  solved,  gives 

x=2, 

and  this  value  of  x,  substituted  in  either  of  the  original  equa- 
tions, gives 

y=3. 
The  method  thus  exemplified  is  expressed  in  the  following 

RULE. 

Find  an  expression  for  the  value  of  one  of  the  unknown  quan- 
tities in  one  of  the  equations  ;  then  substitute  this  value  in  the 
place  of  its  equal  in  the  other  equation. 

ELIMINATION  BY  COMPAEISON. 

(110.)  To  illustrate  this  method,  take  equations  (1.)  of  the 
preceding  Article.  Derive  from  each  equation  an  expression 
for  x  in  terms  of?/,  and  we  shall  have 

x=12-y, 
x=  6+y. 

These  two  values  of  x  must  be  equal  to  each  other,  and  by 
comparing  them  we  shall  obtain 


an  equation  involving  only  one  unknown  quantity  ; 
whence  —%- 


80  SIMPLE    EQUATIONS 

Substituting  this  value  of  y  in  the  expression  #=6+y,  and 
we  find  x  =9,  as  before. 

Again,  take  equations  (2.)  of  the  preceding  Article. 
From  equation  first,  we  find 

13—  2x 

r  75MT' 

and  from  equation  second, 

22  -5x 

y=-r- 

Putting  these  values  of  y  equal  to  each  other,  we  have 


an  equation  containing  only  x,  whence  we  obtain 

x=2. 

Substituting  this  value  of  x  in  either  of  the  preceding  ex- 
pressions for  y,  we  find 

y=3. 
The  method  thus  exemplified  is  expressed  in  the  following 

RULE. 

Find  an  expression  for  the  value  of  the  same  unknown  quan- 
tity in  each  of  the  equations,  and  form  a  new  equation  by  put- 
ting one  of  these  values  equal  to  the  other. 

ELIMINATION  BY  ADDITION  AND  SUBTRACTION. 

(111.)  To  illustrate  this  method,  take  equations  (1.)  of  Art. 
109.  Since  the  coefficients  of  y  in  the  two  equations  are 
equal  and  have  contrary  signs,  we  may  eliminate  this  letter  by 
adding  the  two  equations  together,  whence  we  obtain 

2x=18, 
or  x=  9. 

We  may  now  deduce  the  value  of  y  by  substituting  the 
value  of  x  in  one  of  the  original  equations.  Taking  the  first 
for  example,  we  have 

9+y=12, 
whence  y=  3. 

Since  the  coefficients  of  x  are  equal  in  the  two  original 
equations,  we  might  have  eliminated  this  letter  by  subtracting 


CONTAINING    TWO    Oil    MORE    UNKNOWN     QUANTITIES.  81 

one  equation  from  the  other.     Subtracting  the  first  from  the 
second,  we  obtain 

2y=6, 
or    ?/=3. 

Let  us  apply  the  same  method  to  equations  (2.)  of  Art.  109. 
We  perceive  that  if  we  could  deduce  from  the  proposed  equa- 
tions two  other  equations,  in  which  the  coefficients  of  y  should 
be  equal,  the  elimination  of  y  might  be  effected  by  subtracting 
one  of  these  new  equations  from  the  other. 

It  is  easily  seen  that  we  shall  obtain  two  equations  of  the 
form  required,  if  we  multiply  all  the  terms  of  each  equation  by 
the  coefficient  of  y  in  the  other.  Multiplying,  therefore,  all 
the  terms  of  equation  first  by  4,  and  all  the  terms  of  equation 
second  by  3,  they  become 

8x+12y=52, 


Subtracting  the  former  of  these  equations  from  the  latter,  we 
find 

7x=l4, 
whence  x=  2. 

In  like  manner,  in  order  to  eliminate  x,  multiply  the  first  of 
the  proposed  equations  by  5,  and  the  second  by  2,  they  will 
become 


8y=44. 

Subtracting  the  latter  of  these  two  equations  from  the  for- 
mer, we  have 

7y=21, 
whence  y=  3. 

This  last  method  is  expressed  in  the  following 

RULE. 

Multiply  or  divide  the  equations,  if  necessary,  in  such  a  man- 
ner, that  one  of  the  unknown  quantities  shall  have  the  same  coef- 
ficient in  both.  Then  subtract  one  equation  from  the  other,  if 
the  signs  of  these  coefficients  are  the  same,  or  add  them  together 
if  the  signs  are  different. 

F 


82  SIMPLE    EQUATIONS 


EXAMPLES. 


(112.)  Ex.  1.  Given  5z+4y=58  )  to  find  the  values  of  x 

—  67  \ 


and  y. 

By  the  first  method. 

From  the  second  equation  we  find 
3z=67-7y. 

Therefore  x=  —  -  —  . 

o 

Substituting  this  value  of  x  in  the  first  equation, 


Hence  335-35y+12y=174. 

By  transposition,       335  —  174=35y—   12y, 
or  161=23?/. 

Therefore  y—^> 

Substituting  this  value  of  y  in  the  expression  for  the  value 
of  x  given  above,  it  becomes 

67-7X7_67-49_18_ 

~3~          ~3~   ~T 
Thus  we  have          y=I7,  and  x=6. 

By  the  second  method. 

From  the  first  equation  we  find 

5x=58—4y, 

58  -4y 
whence  x—  —  r  —  . 

D 

67-  7y 
From  the  second  equation,  x—  —  TT-^-* 

o 

58-4v     67—  7y 

Therefore  —  —  ^-=  —  --^. 

5  o 

Clearing  of  fractions,  174—  12?/=335-35y. 
By  transposition,         35y—  12^=335—  174, 
or  23y=161. 

Therefore  y=rl> 

whence,  as  before,  .77=6. 


CONTAINING    TWO    OR    MORE    UNKNOWN    QUANTITIES.  83 

By  the  third  method. 

Multiplying  the  second  equation  by  5  and  the  first  by  3,  we 
obtain 


and  15z+12?/=174. 

By  subtraction,  23?/=161, 

or  y=     7. 

Whence,  from  equation  first, 

5z=58-4y=58—  28=30, 
and  therefore  x  =6. 

Thus  the  same  example  may  be  solved  by  either  of  the  three 
methods,  and  each  method  has  its  advantages  in  particular 
cases.  Generally,  however,  the  first  two  methods  give*  rise  to 
fractional  expressions  which  occasion  inconvenience  in  prac- 
tice, while  the  third  method  is  not  liable  to  this  objection. 
When  the  coefficient  of  one  of  the  unknown  quantities  in  one 
of  the  equations  is  equal  to  unity,  this  inconvenience  does  not 
occur,  and  the  method  by  substitution  may  be  preferable  ;  the 
third  will,  however,  commonly  be  found  most  convenient. 

Ex.  2.  Given  llx+3y=WO  \  . 

.       „  _     4  1  to  find  the  values  of  x  and  y. 

Multiplying  the  first  equation  by  7  and  the  second  by  3,  we 
obtain 

77z+21y=700, 
I2x-2ly=   12. 

Therefore,  by  addition,     89^=712, 

or  x=     8. 

From  equation  first,  3y=  100—1  Ix, 

=  100-88=12, 
and  2/=4. 

These  values  of  x  and  y  may  be  easily  verified  by  substitu- 
tion in  the  original  equations. 

Thus,         11X8+3X4=100;  or  88  +  12=100. 
And  4X8—7X4=     4;  or  32-28=     4. 


Ex.  3.  Given  2+|=7  I 

^  to  find  the  values  of  x  and  y. 

ZC       *J  i 

o       o 

.Arcs.  x=6,  v=12. 


84  SIMPLE    EQUATIONS 

x+2 

4-  wy=ai     | 

to  find  the  values  of  x  and  y. 


Ex.  4.  Given  ^+  8y=31   1 


^+10*=192J 


x+3 
Ex.  5.  Given  2y-—  =7  find 


.  6.  Given  -+-=m 

^        ^  to  find  the  values  of  #  and  y. 
d  y 


c 


be—  ad  be—  ad 

Ans.  x=—r  -  -3;  y=  --  . 
no—  ma  mc—na 

(113.)  When  a  problem  involves  a  large  number  of  quanti- 
ties, it  is  common  to  designate  a  part  of  them  by  different  let 
ters,  and  for  the  remaining  quantities  to  employ  the  same  let 
ters  accented  or  numbered. 

Thus,  a,  a',     a",    a'",  a""  .     %.  .  f,f  ,  '  :V)    v'.     a<"> 

«(1),  a(3),   a(3),  «(4)  .'      r-,  •         •         •     «(>n 


are  used  to  denote  different  quantities,  though  they  generally 
imply  some  connection  between  the  quantities  which  they  rep- 
resent. a1  is  read  a  prime;  a",  a  second;  a"1,  a  third,  &c. 
We  must  carefully  distinguish  between  a.2  and  a2;  -between  a4 
and  a\  &c.  In  the  one  case,  the  numerals  are  exponents,  an  ' 
denote  powers  of  a  ;  while  in  the  other  case,  the  numerals  are 
only  used  for  the  sake  of  convenience  to  denote  distinct  quan- 
tities. Examples  showing  the  convenience  of  this  notation  will 
be  found  in  Sections  XIX.  and  XX. 

Ex.  7.  Given  ax  +by  =c  ( 

,    ,  7  ,  _  /  C  to  find  the  values  of  x  and  y. 

b'c—bc'  ac'—a'c 

Ans.  X——T-.  -  77,*  <y=—T~,  -  n> 

ab'—a'b    y     ab'—a'b 

The  symmetry  of  these  expressions  is  well  calculated  to  fix 
them  in  the  memory. 

Ex.  8.  What  fraction  is  that,  to  the  numerator  of  which,  if  4 


CONTAINING    TWO    OR    MORE    UNKNOWN    QUANTITIES.  85 

be  added,  the  value  is  one  half;  but  if  7  be  added  to  the  de- 
nominator, its  value  is  one  fifth  ? 

Let  -  represent  the  fraction  required. 

Then,  by  the  first  condition, 

z+4     1 

-  =-;  whence  2x+S=y. 

By  the  second  condition, 

=-  ;  whence  &c=y+7. 


Subtracting  the  first  equation  from  the  second,  we  have 

3z-8=7, 

whence  3#=15, 

or  x=5. 

Therefore,  y=2x+8=W+8=lQ, 

and  the  fraction  is  T5¥. 

P      f  5+41 

Pro°f-  -w=* 

5         1 

18+7=5' 

Ex.  9.  A  certain  sum  of  money,  put  out  at  simple  interest, 
amounts  in  8  months  to  81488,  and  in  15  months  it  amounts  to 
81530.  What  is  the  sum  and  rate  per  cent.  ? 

Ex.  10.  A  sum  of  money  put  out  at  simple  interest  amounts 
in  m  months  to  a  dollars,  and  in  n  months  to  b  dollars. 

Required  the  sum  and  rate  per  cent.  ? 

,    na—mb  b—a 

Ans.  The  sum  is  ---  ;  the  rate  is  1200X 


n— m  na—mb' 

Ex.  11.  There  is  a  number  consisting  of  two  digits,  the 
second  of  which  is  greater  than  the  first;  and  if  the  number 
be  divided  by  the  sum  of  its  digits,  the  quotient  is  4 ;  but  if 
the  digits  be  inverted,  and  that  number  be  divided  by  a  num- 
ber greater  by  two  than  the  difference  of  the  digits,  the  quo- 
tient is  14. 

Required  the  number. 

Let  x  represent  the  left  hand  digit, 
and      y  "  right  hand  digit. 

Then,  since  x  stands  in  the  place  of  tens,  the  number  will  be 
represented  by  IQx+y. 


fiii  ....;  SIMPLE    EdUATIONS 

Hence,  by  the  first  condition, 

Wx+y_ 
x+y  = 

By  the  second  condition, 

Wy+x__ 
y-X+2~ 

Whence  x=4,  y=8, 

and  the  required  number  is  48. 

Ex.  12.  A  boy  expends  thirty  pence  in  apples  and  pears, 
buying  his  apples  at  4  and  his  pears  at  5  for  a  penny,  and 
afterward  accommodates  his  friend  with  half  his  apples  and 
one  third  of  his  pears  for  13  pence.  How  many  did  he  buy 
of  each  ? 

Ex.  13.  A  father  leaves  a  sum  of  money  to  be  divided  among 
his  children,  as  follows :  the  first  is  to  receive  $300  and  the 
sixth  part  of  the  remainder ;  the  second  8600  and  the  sixth 
part  of  the  remainder ;  and,  generally,  each  succeeding  one 
receives  8300  more  than  the  one  immediately  preceding,  to- 
gether with  the  sixth  part  of  what  remains.  At  last  it  is  found 
that  all  the  children  receive  the  same  sum.  What  was  the 
fortune  left  and  the  number  of  children  ? 

Ans.  The  fortune  was  $7500,  the  number  of  children  5. 

Ex.  14.  A  sum  of  money  is  to  be  divided  among  several 
persons,  as  follows :  the  first  receives  a  dollars  together  with 
the  rath  part  of  the  remainder ;  the  second  2a  together  with 
the  nth  part  of  the  remainder;  and  each  succeeding  one  a  dol- 
lars more  than  the  preceding,  together  with  the  nth  part  of 
the  remainder ;  and  it  is  found,  at  last,  that  all  have  received 
the  same  sum.  What  was  the  amount  divided,  and  the  num- 
ber of  persons  ? 

Ans.  The  amount  =a(n—  I)2,  the  number  of  persons  —n—  I. 

EdUATIpNS  WHICH  CONTAIN  THREE  OB  MORE  UNKNOWN 
dUANTITIES.  _,.r, 

(114.)  Let  us  now  consider  the  case  of  three  equations  in- 
volving three  unknown  quantities. 

Take  the  system  of  equations, 

3x+2y+  z=I6,  (1.) 

2x+2y+2z=l8,  (2.) 

2x+3y+  z  =  l7.  (3.) 


CONTAINING    TWO    OR    MORE    UNKNOWN    QUANTITIES.  87 

In  order  to  eliminate  z  between  equations  (1.)  and  (2.),  we 
will  divide  both  members  of  the  second  equation  by  two  ;  we 
thus  obtain 


Subtracting  this  from  the  first  equation,  we  find  a  new  equa- 
tion containing  but  two  unknown  quantities, 

2x+y=1.  (a.) 

In  order  to  eliminate  z  between  equations  (1.)  and  (3.),  sub- 
tract the  former  from  the  latter,  which  gives 

-x+y=l.    '  ((3.) 

From  the  two  equations  (a.)  and  ((3.),  one  may  be  deduced 
containing  only  one  unknown  quantity.  For,  by  subtracting 
the  one  from  the  other,  we  have 

3x=6,  or  x=2. 
Substituting  this  value  of  x  in  equation  (0.),  we  obtain 

y=3. 

Substituting  these  values  of  x  and  y  in  equation  (1.),  we  ob- 
tain 

3X2+2X3+z=16. 
Hence  z—  4. 

These  values  of  x,  y,  and  z  may  be  verified  by  substitution 
in  the  original  equations. 

We  have  effected  the  elimination  in  this  case  by  method 
third,  Art.  Ill;  but  either  of  the  other  methods  might  have 
been  employed.  Hence,  to  solve  three  equations  containing 
three  unknown  quantities,  we  have  the  following 

RULE. 

(115.)  From  the  three  equations,  deduce  two  containing  only 
two  unknown  quantities  ;  then  from  these  two  deduce  one  con- 
taining only  one  unknown  quantity. 

Ex.  15.  Given    x+  y+  z=29  (1.)  > 

x+2y+3z=62  (2.)  >  to  find  x,  y,  and  z. 

i*+iy+i*=io  (3.)  ) 

Subtract  equation  (1.)  from  (2.),  and  we  obtain 

y+2z=33;  (a.) 

clearing  equation  (3.)  of  fractions,  we  have 

6x+4y+3z=120.  (4.) 

5 


88  SIMPLE    EQUATIONS 

Multiplying  equation  (1.)  by  6, 

6z+6y+6z=174.  (5.) 

Subtracting  (4.)  from  (5.),  2y+3z=54.  (0.) 

We  have  thus  obtained  two  equations,  (a.)  and  (0.),  contain- 
ing two  unknown  quantities. 

Multiplying  (a.)  by  2,  we  have  2y+4z=66,  (6.) 

Subtracting  (|3.)  from  (6.),  z=l2. 
Substituting  this  value  of  %  in  (0.),  we  obtain 

2y+36=54. 

Whence  y=$> 

Substituting  these  values  of  y  and  z  in  equation  (1.), 


Whence  x=S. 

These  values  may  be  verified  as  in  former  examples. 

Ex.  16.  Given  2x+4y-3z=22  \ 

4x—2y+5z=l8  >  to  find  x,  y,  and  z. 


Ans.  # 
.  17.  Given  x+y—a  \ 

x+  z=b  >  to  find  x,  y,  and  z. 


18.  Given 

=15  >  to  find  x,  y,  and  z. 


(116.)  If  we  had  /owr  equations  involving  four  unknown 
quantities,  we  might,  by  the  methods  already  explained,  elim- 
inate one  of  the  unknown  quantities.  We  should  thus  obtain 
three  equations  between  three  unknown  quantities,  which  might 
be  solved  according  to  Art.  114.  So,  also,  if  we  had  Jive 
equations  involving  Jive  unknown  quantities,  we  might,  by  the 
same  process,  reduce  them  to  four  equations  involving  four 
unknown  quantities  ;  then  to  three,  and  so  on.  By  following 
the  same  method,  we  might  resolve  a  system  of  any  number 
of  equations  of  the  first  degree.  Hence,  if  we  have  m  equa- 
tions involving  m  unknown  quantities,  we  proceed  by  the  fol- 
lowing 


RULE. 


1.  Combine  successively  any  one  of  the  equations  with  each 
of  the  others,  so  as  to  eliminate  the  same  unknown  quantity  ;  we 


CONTAINING    TWO    OR    MORE    UNKNOWN    QUANTITIES.  69' 

thus  obtain   m—l  new  equations  containing  m— 1   unknown 


2.  Eliminate  another  unknown  quantity  by  combining  any 
one  of  these  new  equations  with  the  others;  there  will  result 
m— 2  equations  containing  m— 2  unknown  quantities. 

3.  Continue  this  series  of  operations  until  there  results  a 
single  equation   containing  but  one  unknown  quantity,  from 
which  the  value  of  this  unknown  quantity  is  easily  deduced. 
Then  by  going  back,  step  by  step,  to  one  of  the  original  equa- 
tions, the  values  of  the  other  unknown  quantities  may  be  suc- 
cessively determined. 

Ex.  19.  Given  7x-2z  +3^= 

4y— 2z  +    t=ll 

5y—3x—2u=  8  )>  to  find  x,  y,  z,  u,  and  t. 
4y—3u+2  t=  9 
3z+8u=33j 

Ans.  x=2,  y=4,  z=3,  u=3,  t=l. 

Either  of  the  unknown  quantities  may  be  selected  as  the 
one  to  be  first  exterminated.  It  is,  however,  generally  best  to 
begin  with  that  which  has  the  smallest  coefficients  ;  and  if  each 
of  the  unknown  quantities  is  not  contained  in  all  the  proposed 
equations,  it  is  generally  best  to  begin  with  that  which  is  found 
in  the  least  number  of  equations. 

Ex.  20.  A  person  owes  a  certain  sum  to  two  creditors.  A* 
one  time  he  pays  them  8530,  giving  to  one  four  elevenths  of 
the  sum  which  is  due,  and  to  the  other  $ 30  more  than  one 
sixth  of  his  debt  to  him.  At  a  second  time  he  pays  them  $420, 
giving  to  the  first  three  sevenths  of  what  remains  due  to  him, 
and  to  the  other  one  third  of  what  remains  due  to  him.  What 
were  the  debts  ? 

Ex.  21.  If  A  and  B  together  can  perform  a  piece  of  work 
in  12  days,  A  and  C  together  in  15  days,  and  B  and  C  in  20 
days,  how  many  days  will  it  take  each  person  to  perform  the 
same  work  alone  ? 

This  Problem  is  readily  solved  by  first  finding  in  what  time 
they  could  finish  it  if  all  worked  together. 

Ex.  22.  If  A  and  B  together  can  perform  a  piece  of  work 
in  a  days,  A  and  C  together  in  b  days,  and  B  and  C  in  c  days, 


90  SIMPLE    EQUATIONS 


how  many  days  will  it  take  each  person  to  perform  the  same 
work  alone  ? 

2abc 
Ans.  A  requires      --      days, 


2abc 

B        "          ,  ,  ,  -  days, 
ab+bc—ac      J 


days. 

ab+ac—bc      J 

(11*7.)  Hitherto  we  have  supposed  the  number  of  equations 
equal  to  the  number  of  symbols  employed  to  denote  the  un- 
known quantities.  This  must  be  the  case  with  every  problem, 
in  order  that  it  may  be  determinate  ;  that  is,  that  it  may  not 
admit  of  an  indefinite  number  of  solutions. 

Suppose,  for  example,  that  a  problem  involving  two  un- 
known quantities  (x  and  y)  leads  to  the  single  equation 

x—  y=3. 

Now  if  we  make  y=l,  then  x=4  ; 

y=2,  then  #=5; 

y=3,  then  x=Q  ; 

y=4,  then  #=7, 

&c.,  &c.  ; 

and  each  of  these  systems  of  values,  1  and  4,  2  and  5,  3  and  6, 
&c.,  substituted  for  x  and  y  in  the  original  equation,  will  sat- 
isfy it  equally  well.  Hence  the  problem  is  indeterminate  ;  that 
is,  admits  of  an  indefinite  number  of  solutions. 

(118.)  If  we  had  two  equations  involving  three  unknown 
quantities,  we  could,  in  the  first  place,  eliminate  one  of  the  un- 
known quantities  by  means  of  the  proposed  equations,  and 
thus  obtain  one  equation  containing  two  unknown  quantities, 
which  would  be  satisfied  by  an  infinite  number  of  systems  of 
values.  Therefore,  in  order  that  a  problem  may  be  determ- 
inate, its  enunciation  must  contain  as  many  different  condi- 
tions as  there  are  unknown  quantities,  and  each  of  these  con- 
ditions must  be  expressed  by  an  independent  equation. 

Equations  are  said  to  be  independent  when  they  express 
conditions  essentially  different  ;  and  dependent  when  they  ex- 
press the  same  conditions  under  different  forms. 

Thus,    x+y=  7  ) 

2x+v=lO  \  are  independent  equations. 


CONTAINING    TWO    OB    MORE    UNKNOWN    QUANTITIES.  91 

But    x+  y=  7 


=  7  ) 

=  14  J  are  not  Dependent, 


because  the  one  may  be  deduced  from  the  other. 

(119.)  If,  on  the  contrary,  the  number  of  independent  equa- 
tions exceeds  the  number  of  unknown  quantities,  these  equa- 
tions will  be  contradictory. 

For  example,  let  it  be  required  to  find  two  numbers  such 
that  their  .sum  shall  be  7,  their  difference  1,  and  their  product 
100. 

From  these  conditions  we  derive  the  following  equations  : 
x+y=     7, 
x-y=     1, 
xy=WO. 
From  the  first  two  equations  we  easily  find 

x=4,  and  y=S. 

Hence  the  third  condition,  which  requires  that  their  product 
ihall  be  equal  to  100,  can  not  be  fulfilled. 


SECTION  IX. 


DISCUSSION  OF  EQUATIONS  OF  THE 
FIRST  DEGREE.    INEQUALITIES. 

(120.)  To  discuss  a  problem  or  an  equation  is  to  determine 
the  values  which  the  unknown  quantities  assume  for  particular 
hypotheses  made  upon  the  values  of  the  given  quantities,  and 
to  interpret  the  peculiar  results  obtained.  The  term,  there- 
fore, is  not  strictly  applicable,  except  to  problems  which  are 
stated  in  the  most  general  form,  like  some  of  those  in  Arts.  106 
and  107.  If  the  sum  of  two  numbers  is  represented  by  a  and 
their  difference  by  &,  the  greater  number  will  be  expressed  by 

a    b  .        .      a     b 

-+-,  and  the  less  by  5— •-•     -"-ere  a  and  b  may  have  any 

values  whatever,  and  still  these  formula  will  always  hold  true. 
It  frequently  happens  that,  by  .attributing  different  values  to  the 
letters  which  represent  known  quantities,  the  values  of  the  un- 
known quantities  assume  peculiar  forms  which  deserve  con- 
sideration. 

(121.)  We  may  obtain  five  species  of  values  for  the  unknown 
qi  lantity  in  a  problem  of  the  first  degree. 
I.  Positive  values. 

II.  Negative  values. 

III.  Values  of  the  form  of  zero,  or  -r. 

A 

IV.  Values  of  the  form  of  — . 

V.  Values  of  the  form  of  -. 
We  will  consider  these  five  cases  in  succession. 


DISCUSSION  OF  EQUATIONS,   ETC.  93 

I.  Positive  values  are  generally  answers  to  problems  in  the 
sense  in  which  they  are  proposed.  Nevertheless,  all  positive 
values  will  not  always  satisfy  the  enunciation  of  a  problem. 
If,  for  example,  a  problem  requires  an  answer  in  whole  num- 
bers, and  we  obtain  a  fractional  value,  the  problem  is  impossi- 
ble. Thus,  in  Problem  17,  page  71,  it  is  implied  that  the  value 
of  x  must  be  a  whole  number,  although  this  condition  is  not 
expressed  in  the  equations.  It  would  be  easy  to  change  the 
data  of  the  problem  so  as  to  obtain  a  fractional  value  of  x, 
which  would  indicate  an  impossibility  in  the  problem  pro- 
posed. Problem  43,  page  76,  is  of  the  same  kind ;  also  Ex. 
11,  page  85. 

If  the  value  obtained  for  the  unknown  quantity,  even  when 
positive,  does  not  satisfy  all  the  conditions  of  the  problem,  the 
problem  is  impossible  in  the  form  proposed. 

(122.)  II.  Negative  values. 

Let  it  be  proposed  to  find  a  number,  which,  added  to  the 
number  6,  gives  for  a  sum  the  number  a. 

Let  x  =  the  required  number. 

Then,  by  the  terms  of  the  problem, 

x+b=a,  whence  x=a—b. 

This  formula  will  give  the  value  of  x  for  every  case  of  the 
proposed  problem. 

For  example,  let         «=7,  and  b=4. 

Then  x=l  —  4=3. 

Again,  let  a =  5,  and  b=S. 

Then  x=5_8=-3. 

We  thus  obtain  for  x  a  negative  value.  How  is  it  to  be  in- 
terpreted ? 

By  referring  to  the  problem,  we  see  that  it  is  proposed  to 
find  a  number  which,  added  to  8,  shall  make  it  equal  to  5. 
Considered  arithmetically,  the  problem  is  plainly  impossible. 
Nevertheless,  if  in  the  equation  8+x=5,  we  substitute  for  +z 
its  value  —3,  it  becomes 

8-3=5, 
an  identical  equation ;  that  is,  8  diminished  by  3  is  equal  to  5. 

The  negative  solution  x=  —  3,  shows,  therefore,  the  impossi- 
bility of  satisfying  the  enunciation  of  the  problem  as  above 
stated  ;  but,  taking  this  value  of  a:  with  a  contrary  sign,  we  see 
that  it  satisfies  the  enunciation  when  modified  as  follows : 


94  DISCUSSION  or  EQUATIONS 

To  find  a  number  which,  subtracted  from  8,  gives  a  differ- 
ence of  5 ;  an  enunciation  which  differs  from  the  former  only 
in  this,  that  we  put  subtract  for  add,  and  difference  for  sum. 

If  we  wish  to  solve  this  new  question  directly,  we  shall 
have 

8— x—5. 

Whence  x =8— 5,  or  x=3. 

(123.)  For  another  example,  take  Problem  50,  page  77. 
The  age  of  the  father  being  represented  by  a,  and  that  of  the 

son  by  b  ;  then  — — y-  will  represent  the  number  of  years  be- 
fore the  age  of  the  father  will  be  n  times  that  of  the  son. 
Thus,  suppose      #=54,  &=9,  and  ?i=4. 

54-36     18 
Then  ,=___=_=«, 

That  is,  the  father  having  lived  54  years  and  the  son  9,  in  6 
years  more  the  father  will  be  60  years  old  and  the  son  15. 
But  60  is  4  times  15  ;  hence  this  value,  x=6j  satisfies  the  enun- 
ciation of  the  problem. 

Again,  suppose  #=45,  &=15,  and  n=4. 

45-60     -15 
Then  x= — - — =—-—=  — 5. 

Here  again  we  obtain  a  negative  solution.  How  are  we  to 
interpret  it  ? 

By  referring  to  the  problem,  we  see  that  the  age  of  the  son 
is  already  more  than  one  fourth  that  of  the  father,  so  that  the 
time  required  is  already  past  by  five  years.  The  value  of  x 
just  obtained,  taken  with  a  contrary  sign,  satisfies  the  following 
enunciation : 

A  father  is  45  years  old,  his  son  15 ;  how  many  years  since 
the  age  of  the  father  was  four  times  that  of  his  son  ? 

The  equation  corresponding  to  this  new  enunciation  is 

__    _45~x 

Whence  60— 4x=45— x;  and  x=5. 

(124.)  Reasoning  from  analogy,  we  deduce  the  following 
general  principles : 

1.  Every  negative  value  found  for  the  unknown  quantity  in  a 


OF    THE    FIRST    DEGREE.  95 

problem  of  the  first  degree,  indicates  an  absurdity  in  the  condi- 
tions of  the  problem,  or  at  least  in  its  algebraic  statement. 

2.  This  value,  taken  with  a  contrary  sign,  may  be  regarded 
as  the  answer  to  a  problem',  whose  enunciation  only  differs  from 
that  of  the  proposed  problem  in  this,  that  certain  quantities 
which  were  ADDED  should  have  been  SUBTRACTED,  and  recipro- 
cally. 

(125.)  In  what  case  would  the  value  of  the  unknown  quan- 
tity in  Prob.  20,  page  72,  be  negative  ? 

Ans.  When 

Thus,  let       m=20,  n=25,  and  a=6Q  miles. 

Then        *==-18- 


To  interpret  this  result,  observe  that  it  is  impossible  that  the 
second  train,  which  moves  the  slowest,  should  overtake  the 
first.  At  the  time  of  starting,  the  distance  between  them  was 
60  miles,  and  every  subsequent  hour  the  distance  increases. 
If,  however,  we  suppose  the  two  trains  to  have  been  moving 
uniformly  along  an  endless  road,  it  is  obvious  that  at  some 
former  time  they  must  have  been  together. 

This  negative  solution  then  shows  an  absurdity  in  the  con- 
ditions of  the  problem.  The  problem  should  have  been  stated 
thus: 

Two  trains  of  cars,  60  miles  apart,  are  moving  in  the  same 
direction,  the  forward  one  25  miles  per  hour,  the  other  20. 
How  long  since  they  were  together  ? 

To  solve  this  problem,  let  x  =  the  required  number  of  hours. 

Then  25x  —  the  distance  traveled  by  the  first  train, 

20;e  =  "  "         second  train. 

And  since  they  are  now  60  miles  apart, 


Hence  5z=60, 

and  2;=  +  12. 

We  thus  obtain  a  positive  value  of  x. 

In  order  to  include  both  of  these  cases  in  the  same  enuncia- 
tion, the  question  should  have  been  asked,  Required  the  time  of 
'heir  being  together,  leaving  it  uncertain  whether  the  time  was 
vast  or  future. 

In  what  case  would  the  value  of  one  of  the  unknown  qnan 


96  DISCUSSION   OF   EQUATIONS 

titles  in  Problem  34,  page  74,  be  negative  ?  Why  should  it  be 
negative  ?  and  how  could  the  enunciation  be  corrected  for  this 
case? 

In  what  case  would  the  value  of  one  of  the  unknown  quan- 
tities in  Problem  4,  page  67,  be  negative  ? 

0 
(126.)  III.  Values  of  the  form  of  zero,  or  -T. 

In  what  case  would  the  value  of  the  unknown  quantity  in 
Problem  20,  page  72,  become  zero,  and  what  would  this  value 
signify  ? 

Ans.  This  value  becomes  zero  when  a=0,  which  signifies 
that  the  two  trains  are  together  at  the  outset. 

In  what  case  would  the  value  of  the  unknown  quantity  in 
Problem  50,  page  77,  become  zero,  and  what  would  this  value 
signify  ? 

Ans.  When  a=nb,  which  signifies  that  the  age  of  the  fa- 
ther is  now  n  times  that  of  the  son. 

In  what  case  would  the  values  of  the  unknown  quantities  in 
Problem  38,  page  75,  become  zero,  and  what  would  this  sig- 
nify? 

When  a  problem  gives  zero  for  the  value  of  the  unknown 
quantity,  this  value  is  sometimes  applicable  to  the  problem, 
and  sometimes  it  indicates  an  impossibility  in  the  proposed 
question. 

(127.)  IV.  Values  of  the  form  of-. 

In  what  case  does  the  value  of  the  unknown  quantity  in 
oblem  20,  pa 

pret  this  result  ? 

Ans.  When  m=n. 

On  referring  to  the  enunciation  of  the  problem,  we  see  that 
it  is  absolutely  impossible  to  satisfy  it  ;  that  is,  there  can  be 
no  point  of  meeting,  for  the  two  trains  being  separated  by  the 
distance  a,  and  moving  equally  fast,  will  always  continue  at 

the  same  distance  from  each  other.  The  result  -  may  then 
be  regarded  as  indicating  an  impossibility. 


Problem  20,  page  72,  reduce  to  —  ?  and  how  shall  we  inter 


OF    THE    FIRST    DEGREE.  97 

The  symbol  —  is  sometimes  employed  to  represent  infinity  ; 

and  for  the  following  reason  : 

When  the  difference  m— n,  without  being  absolutely  nothing, 

is  very  small,  the  quotient     _     is  very  large. 
For  example,  let  m— n=0.01. 

Then  x=— —=-^-= 

m—n    .01 

Let  m— n=0.0001, 

a  a 


m—n     .0001 

Hence,  if  the  difference  in  the  rates  of  motion  is  not  zero, 
the  two  trains  must  meet,  and  the  time  will  become  greater 
and  greater  as  this  difference  is  diminished.  If,  then,  we  sup- 
pose this  difference  less  than  any.  assignable  quantity,  the  time 

represented  by  -     -  will  be  greater  than  any  assignable  quan- 
tity, or  infinite. 

j^ 
Hence  we  infer,  that  every  expression  of  the  form  — ,  found 

for  the  unknown  quantity,  indicates  the  impossibility  of  satis- 
fying the  problem,  at  least  in  finite  numbers. 

In  what  case  would  the  value  of  the  unknown  quantity  m 

^ 
Problem  10,  page  70,  reduce  to  the  form  —  ?  and  how  shall 

we  interpret  this  result  ? 

(128.)  V.  Values  of  the  form  of-. 

In  what  case  does  the  value  of  the  unknown  quantity  in 
Problem  20,  page  72,  reduce  to  -  ?  and  how  shall  we  interpret 

this  result? 

Ans.  When  a=0,  and  m=n. 

To  interpret  this  result,  let  us  recur  to  the  enunciation,  and 
observe  that,  since  a  is  zero,  both  trains  start  from  the  same 
point ;  and  since  they  both  travel  at  the  same  rate,  they  will 
always  remain  together,  and  therefore  the  required  point  of 
meeting  will  be  any  where  in  the  road  traveled  over.  Th 

G 


98  OF    ZERO    AND    INFINITY. 

problem,  then,  is  entirely  indeterminate,  or  admits  of  an  infinite 
number  of  solutions,  and  the  expression  -  may  represent  any 
finite  quantity. 

We  infer,  therefore,  that  an  expression  of  the  form  -  found 

for  the  unknown  quantity,  generally  indicates  that  it  may  have 
any  value  whatever.  In  some  cases,  however,  this  value  is 
subject  to  limitations. 

In  what  case  would  the  values  of  the  unknown  quantities  in 

Problem  44,  page  76,  reduce  to  -  ?  and  how  would  they  satisfy 

the  conditions  of  the  problem? 

Ans.  When  a=b=c, 

which  indicates  that  the  coins  are  all  of  the  same  value.     B 
might  therefore  be  paid  in  either  kind  of  coin ;  but  there  is  a 
limitation,  viz.,  that  the  value  of  the  coins  must  be  one  dollar. 
In  what  case  do  the  values  of  the  unknown  quantities  in 

Problem  38,  page  75,  reduce  to  -  ?  and  how  shall  we  interpret 
this  result  ? 


OF  ZERO  AND  INFINITY. 

(129.)  From  Art.  127,  it  is  seen  that  in  Algebra  we  some- 
times have  occasion  to  consider  infinite  quantities.  It  is  nec- 
essary, therefore,  to  establish  some  general  principles  respect- 
ing them. 

An  infinite  quantity  is  one  which  exceeds  any  assignable  limit. 
It  is  often  expressed  by  the  character  ce  .  Thus,  a  line  pro- 
duced beyond  any  assignable  limit  is  said  to  be  of  infinite 
length.  A  surface  indefinitely  extended,  and  also  a  solid  of 
indefinite  extent  in  any  one  of  its  three  dimensions,  are  ex- 
amples of  infinity. 

An  infinite  quantity  does  not  mean  an  infinite  number  of 
terms.  Thus,  the  fraction  |  reduced  to  a  decimal,  is  .333333, 
&c.,  without  end,  but  the  value  of  this  series  is  less  than 
unity. 

Infinite  quantities  are  not  all  equal  among  themselves. 


OF    ZEIiO    AND    INFINITY.  99 

Thus  the  series  1  +  1  +  1  +  1  +  1 +,  &c., 
2+2+2+2+2+,  &c., 
3+3+3+3+3+,  &c., 

continued  to  an  infinite  number  of  terms,  will  each  be  infinite, 
although  the  second  series  will  be  double,  and  the  third  treble 
the  first. 

So,  also,  a  line  may  be  infinitely  extended  both  ways ;  or  it 
may  be  infinitely  extended  in  one  direction,  and  limited  in  the 
other.  In  either  case,  the  line  is  said  to  be  infinite. 

A  quantity  less  than  any  assignable  quantity  is  called  an  in- 
finitesimal, and  is  sometimes  represented  by  0. 

Thus,  take  the  series  of  fractions  TV,  T^¥,  r^>  Toio  <r»  &c. 
By  increasing  the  denominator,  we  diminish  the  value  of  the 
fraction  ;  and  if  the  denominator  be  made  infinitely  great,  the 
quotient  will  be  infinitely  small. 

(130.)  We  have  seen,  in  Art.  127,  that  7^=00 ,  where  a  may 

represent  any  finite  quantity.     That  is, 

If  a  finite  quantity  be  divided  by  zero,  the  quotient  is  infinite. 

From  the  same  equation  we  deduce  — =0.     That  is, 

00 

If  a  finite  quantity  be  divided  by  infinity,  the  quotient  is  zero. 

From  the  same  equation  we  deduce  a=OX  oo  .     That  is, 

If  zero  be  multiplied  by  infinity,  the  product  is  a  finite  quan- 
tity. 

If  a  finite  quantity  be  multiplied  by  a  proper  fraction,  it  will 
be  diminished,  and  the  smaller  the  multiplier,  the  less  the  prod- 
uct. Hence,  if  the  multiplier  be  infinitely  small,  the  product 
will  be  infinitely  small,  or  «XO=0.  That  is, 

If  a  finite  quantity  be  multiplied  by  zero,  the  product  will  be 
zero. 

From  this  equation  we  deduce  CL=-  ;  that  is, 

If  zero  be  divided  by  zero,  the  quotient  may  be  any  finite 
quantity. 

The  greater  the  multiplier,  the  greater  will  be  the  product. 
Hence,  if  a  finite  quantity  be  multiplied  by  infinity,  the  product 
will  be  infinite ;  that  is, 

=  00  . 


100  OP   INEQUALITIES. 

00 

From  this  equation  we  deduce  a  = — ;  that  is, 

00 

If  infinity  be  divided  by  infinity,  the  quotient  may  be  any 
finite  quantity. 

An  infinite  quantity  can  not  be  increased  by  the  addition  of 
a  finite  quantity,  or  diminished  by  its  subtraction  ;  that  is, 

QO  ±a  —  <x> . 

So,  also,  a  finite  quantity  is  not  altered  by  the  addition  or 
subtraction  of  zero ;  that  is,  a±0=<z. 


OF  INEQUALITIES. 

(131.)  In  discussing  algebraical  problems,  as  shown  in  Arts. 
120-128,  it  is  frequently  necessary  to  employ  inequalities,  or 
expressions  of  two  quantities  which  are  not  equal  to  each  oth- 
er. Generally,  the  principles  already  established  for  the  trans- 
formation of  equations  are  applicable  to  inequalities  also. 
There  are,  however,  some  important  exceptions  to  be  noted, 
arising  chiefly  from  the  use  of  negative  expressions  as  quan- 
tities. 

Two  inequalities  are  said  to  subsist  in  the  same  sense  when 
the  greater  quantity  stands  at  the  left  in  both,  or  at  the  right 
in  both ;  and  in  a  contrary  sense  when  the  greater  quantity 
stands  at  the  right  in  one,  and  at  the  left  in  the  other. 

Thus,  9>7and7>6. 

As  also  5<8  and  3<4, 

are  inequalities  which  subsist  in  the  same  sense ;  but  the  ine- 
qualities 

10>6  and  3<7, 
subsist  in  a  contrary  sense. 

(132.)  I.  If  we  add  the  same  quantity  to  both  members  of  an 
inequality,  or  subtract  the  same  quantity  from  both  members,  the 
resulting  inequality  will  always  subsist  in  the  same  sense. 

Thus,  8>3. 

Adding  5  to  each  member, 

8+5>3+5; 

and  subtracting  5  from  each  member, 
8-5>3— 5. 

Again,  take  the  inequality  —  3<— 2. 


OF    INEQUALITIES.  101 

Adding  6  to  each  member,  we  have 

-3+6<-2+6,  or  3<4; 

and  subtracting  6  from  each  member, 

-3-6<-2-6,  or  -9<-8. 

The  student  must  here  bear  in  mind  what  was  stated  in  Art. 
47,  of  two  negative  quantities,  that  is  the  least  whose  numer- 
ical value  is  the  greatest. 

This  principle  enables  us  to  transpose  any  term  from  one 
member  of  an  inequality  to  the  other  by  changing  its  sign. 

Thus,  a2+&2>362-2aa. 

Adding  2«2  to  each  member  of  the  inequality,  it  becomes 


Subtracting  62  from  each  member, 


or  3«2>2&a. 

(133.)  II.  If  we  add  together  the  corresponding  members  of 
two  or  more  inequalities  which  subsist  in  the  same  sense,  the  re- 
sulting inequality  will  always  subsist  in  the  same  sense. 

Thus,  5>4 

4>2 


Adding,  we  obtain  16>9. 

III.  But  if  we  subtract  the  corresponding  members  of  two  or 
more  inequalities  which  subsist  in  the  same  sense,  the  resulting 
inequality  will  NOT  ALWAYS  subsist  in  the  same  sense. 

Take  the  inequalities          4<7 

2<3 

Subtracting,  we  have  4—  2<7—  3,  or  2<4, 
where  the  resulting  inequality  subsists  in  the  same  sense. 

But  take  9<10 

and  6<  8. 

Subtracting,  the  result  is  9—  6>  (not  <)  10—8,  or  3>2, 
where  the  resulting  inequality  subsists  in  the  contrary  sense. 

We  should  therefore  avoid  as  much  as  possible  the  use  of 
this  transformation,  or  when  we  employ  it,  determine  in  what 
sense  the  resulting  inequality  subsists. 

(134.)  IV.  If  we  multiply  or  divide  the  two  members  of  an  in- 
equality by  a  positive  number,  the  resulting  inequality  will  sub- 
sist in  the  same  sense. 


102  OF    INECIUALITIES. 

Thus,  if  a  <    b. 

Then  ma<mb. 

a     b 
And  —  <—  . 

mm 

Also,  if  —#>—&. 

Then  —na>—nb. 

And  -a->-b-. 

n        n 

This  principle  will  enable  us  to  clear  an  inequality  of  frac- 
tions.    Thus,  suppose  we  have 

a2-fra      a- 


3a   ' 
Multiplying  both  members  by  Gad,  it  becomes 


V.  If  we  multiply  or  divide  the  two  members  of  an  inequality 
by  a  negative  number,  the  resulting  inequality  will  subsist  in  a 
contrary  sense. 

Take,  for  example,  8>7. 

Multiplying  both  members  by  —3,  we  have  the  opposite  in- 
equality, 

-24<-21. 

So,  also,  15>12. 

Dividing  each  member  by  —3,  we  have 
-5<-4. 

Therefore,  if  we  multiply  or  divide  the  two  members  of  an 
inequality  by  an  algebraic  quantity,  it  is  necessary  to  ascer- 
tain whether  the  multiplier  or  divisor  is  negative,  for  in  this 
case  the  inequality  subsists  in  a  contrary  sense. 

VI.  If  we  change  the  signs  of  both  members  of  an  inequality, 
we  must  reverse  the  sense  of  the  inequality,  for  this  transforma- 
tion is  evidently  the  same  as  multiplying  both  members  by 
-1. 

(135.)  VII.  If  both  members  of  an  inequality  are  positive 
numbers,  we  can  raise  them  to  any  power  without  changing  the 
sense  of  the  inequality. 

Thus,  5>3, 

so  also,  5a>32,  or  25  >  9. 


OF    INEQUALITIES.  103 

And  if  a  >&, 

then  will  an>bn. 

VIII.  If  both  members  of  an  inequality  are  not  positive  num- 
bers, and  they  be  raised  to  any  power,  the  resulting  inequality 
will  not  always  subsist  in  the  same  sense. 

Thus,  — 2<+3, 

gives  (-2)a<32,  or  4<9, 

where  the  resulting  inequality  subsists  in  the  same  sense. 

But  -3>-5, 

gives  (-3)a<(-5)2,  or  9<25, 

where  the  resulting  inequality  subsists  in  a  contrary  sense. 

IX.  In  extracting  the  root  of  both  members  of  an  inequali- 
ty^ it  is  sometimes  necessary  to  reverse  the  sense  of  the  ine- 
quality. 

Thus,  from  9<25, 

by  extracting  the  square  root,  we  obtain 
either  3<5, 

or  -3>-5. 


EXAMPLES. 

1.  Given  7x— 3<25,  to  find  the  limit  of  x. 


2.  Given  2#+-8<6,  to  find  the  limit  of  a:. 

o 


3.  Given  |+|+5+|+-j|-7>9»  to  find  the  limit  of  x- 

bx  ab^\ 

4.  Given  —-\-cx-  «c<-— 

r£  2       " 

dx—bd          X  to  find  the  limits  of  a;. 


Ans. 


Ans.  #6. 


5.  A  man  being  asked  how  many  dollars  he  gave  for  his 
watch,  replied,  If  you  multiply  the  price  by  4,  and  to  the 
product  add  60,  the  sum  will  exceed  256 ;  but  if  you  multiply 
the  price  by  3,  and  from  the  product  subtract  40,  the  re- 


104  OF    INEQUALITIES. 

mainder  will  be  less  than  113.     Required  the  price  of  the 
watch. 

6.  What  number  is  that  whose  half  and  third  part  added 
together  are  less  than  105,  but  its  half  diminished  by  its  fifth 
part  is  greater  than  33  ? 

7.  The  double  of  a  number  diminished  by  6  is  greater  than 
24,  and  triple  the  number  diminished  by  6  is  less  than  double 
the  number  increased  by  10.     Required  the  number. 


SECTION  X. 


INVOLUTION  AND  POWERS. 

(136.)  According  to  Art.  20,  the  products  formed  by  the  suc- 
cessive multiplication  of  the  same  number  by  itself  are  called  the 
powers  of  that  number. 

Thus,  the  first  power  of  3  is  3. 

The  second  power  of  3  is  9,  or  3X3. 

The  fourth  power  of  3  is  81,  or  3X3X3X3, 
&c.,  &c.,  &c. 

According  to  Art.  21,  the  exponent  is  a  number  or  letter  writ- 
ten a  little  above  a  quantity  to  the  right,  and  denotes  the  number 
of  times  that  quantity  enters  as  a  factor  into  a  product. 

Thus,  the  first  power  of  a  is  a1,  where  the  exponent  is  1, 
which,  however,  is  commonly  omitted. 

The  second  power  of  a  is  a  X  a,  or  a3,  where  the  exponent  2 
denotes  that  a  is  taken  twice  as  a  factor  to  produce  the  pow- 
er aa. 

The  third  power  of  a  is  aXaXa,  or  a3,  where  the  exponent 
3  denotes  that  a  is  taken  three  times  as  a  factor  to  produce 
the  power  aaa. 

The  fourth  power  of  a  is  aXaXaXa,  or  a4. 

Also,  the  nth  power  of  a  is  aXaXaXa  .  .  .  repeated 
as  a  factor  n  times,  and  is  written  an. 

Exponents  may  be  applied  to  polynomials  as  well  as  to  -mo- 
nomials. 

Thus  (a+b+c)*  is  the  same  as 

(a+b+c)  x  (a  +  b+c)  x  (a+b+c), 
or  the  third  power  of  the  entire  expression  a+b  +  c. 

(137.)  According  to  the  rule  for  the  multiplication  of  mono- 
mials, Arts.  49  and  50. 


106  INVOLUTION    AND    POWERS. 


(3a62)a=3a&2  X  3ab*=9a*b\ 
So,  also,       (4a*bc*)*=4a*bc3  X  4a*bc3=  16aW. 
Hence  it  appears  that,  in  order  to  square  a  monomial,  we  must 

square  its  coefficient,  and  multiply  the  exponent  of  each  of  the 

letters  by  2. 

EXAMPLES. 

1.  Required  the  square  of7axy. 

Ans.  49aVya. 

2.  Required  the  square  of  lla*bcd*. 

3.  Required  the  square  of  12d*xy. 

4.  Required  the  square  of  I5ab*cx*. 

5.  Required  the  square  of  I8x*yz*. 

'  According  to  Art.  53,  +  multiplied  by  -f,  and  —  multiplied  by 
—  ,  give  +.  Now  the  square  of  any  quantity  being  the  product 
of  that  quantity  by  itself,  it  necessarily  follows  that  whatever 
may  be  the  sign  of  a  monomial,  its  square  must  be  affected  with 
the  sign  +. 

Thus  the  square  of  +3ax  or  of  —  Sax  is  +9aV. 

(138.)  The  method  of  involving  a  quantity  to  any  power,  is 
easily  derived  from  the  preceding  principles. 

Let  it  be  required  to  form  the  fifth  power  of  2aabz. 

According  to  the  rules  for  multiplication, 
2«8&2  *=  2a«b*  X  2a*b* 


Where  we  perceive 

1.  That  the  coefficient  has  been  raised  to  the  fifth  power. 

2.  That  the  exponent  of  each  of  the  letters  has  been  multi 
plied  by  5. 

In  like  manner, 


=27a6b'c*. 

Hence,  to  raise  a  monomial  to  any  power,  we  have  the  fol- 
lowing 


RULE. 


Raise  the  numerical  coefficient  to  the  given  power,  and  multi- 
ply the  exponent  of  each  of  the  letters  by  the  exponent  of  the 
power  required. 


INVOLUTION    AND    POWERS.  107 

EXAMPLES. 

1.  Required  the  fourth  power  of  4<z6V. 

Ans.  256«4&Va 

2.  Required  the  fifth  power  of  3ax3y5. 

3.  Required  the  third  power  of  6xy*z\ 

4.  Required  the  sixth  power  of  2ad*y5v. 

5.  Required  the  seventh  power  of  2a?bc*. 

6.  Required  the  sixth  power  of  5wsxyzz\ 

(139.)  Let  us  now  consider  the  sign  with  which  the  powti 
should  be  affected. 

We  have  seen,  Art.  137,  that  whatever  may  be  the  sign  of  a 
monomial,  its  square  is  always  positive.  It  is  obvious,  from  the 
same  considerations,  that  the  product  of  an  even  number  of 
negative  factors  is  positive,  but  the  product  of  an  odd  number 
of  negative  factors  is  negative. 

Thus,  —  aX-a=+a* 

—  aX—aX—a=  —  a3 

—aX—aX—aX—a=+a* 

—  aX—aX-aX—aX-a=—af> 

&c.,  &c.,  &c. 

The  product  of  several  factors  which  are  all  positive,  is  in- 
variably positive.  Hence, 

Every  EVEN  power  is  positive,  but  an  ODD  power  has  the  same 
sign  as  its  root. 

EXAMPLES. 

1.  Required  the  square  of  —  2x\ 

Ans.  +4z10. 

2.  Required  the  square  of  —  3xn. 

3.  Required  the  cube  of  —  3a3. 

4.  Required  the  fourth  power  of  —  3a?Uh. 

5.  Required  the  fifth  power  of  —  2a3X3x*y. 

(140.)  A  fraction  is  involved  by  involving  both  the  numerator 
and  denominator. 

1.  Thus,  the  square  of  ^  is  ^X^  ;  which,  by  Art.  89,  is 
equal  to  77,  which,  by  Art.  68,  may  be  written  a2&-a. 


108  INVOLUTION    AND    POWERS. 


A         Sa*b'  9       . 

*          '°r 


2.  Required  the  cube  of  — — . 


3.  Required  the  nth  power  of  —  -. 

(141.)  Hence,  expressions  with  negative  exponents  are  in- 
volved by  the  same  rule  as  those  with  positive  exponents. 
Thus,  let  it  be  required  to  find  the  square  of  a~a. 

This  expression  may  be  written  —  ,  which,  raised  to  the 

second  power,  becomes  —  or  a~\  the  same  result  as  would  be 

obtained  by  multiplying  the  exponent  —3  by  2. 

Ex.  1.  Required  the  square  of  3a?b—  *. 

Ex.  2.  Required  the  square  of  7a-*b*cr-*dx-\ 

Ex.  3.  Required  the  cube  of  —  6ab~*dy-\ 

Ex.  4.  Required  the  fourth  power  of  3a~n6. 

Ex.  5.  Required  the  fifth  power  of  —  2ab~  V. 

(142.)  A  polynomial  is  involved  by  multiplying  it  into  itself 
as  many  times  less  one  as  is  denoted  by  the  exponent  of  the 
power. 

Ex.  1.  Required  the  fourth  power  of  a+b. 
a  +b 
a+b 


a*+ab 


>a,  the  second  power  of  a+b. 


a+b    _ 
a*+2a*b+ab* 

+  a*b+2ab*+b* 

(a+by=aa+3a*b+3ab*+ba,  the  third  power. 
a+b  _ 
+  ab* 


(a+by=a'+4a*b+6a*b*+4aba+b*,  the  fourth  power. 
Ex.  2.  Required  the  fourth  power  of  a—  b. 

Ans.  at-4 
Ex.  3.  Required  the  cube  of  2a—  1. 


INVOLUTION    AND    POWERS.  109 

Ex.  4.  Required  the  fourth  power  of  Sa—  h. 

Ex.  5.  Required  the  square  of  a+b+c. 

Hence  it  appears  that  the  square  of  a  trinomial  is  composed 
of  the  sum  of  the  squares  of  all  the  terms,  together  with  twice  the 
sum  of  the  products  of  all  the  terms  multiplied  together  two  and 
two. 

Ex.  6.  Required  the  cube  of  2ab+cd. 

Ex.  7.  Required  the  fourth  power  of  a?+b3. 

Ex.  S.  Required  the  cube  of  a+-. 


Ex.  9.  Required  the  cube  . 

Z 

Ex.  10.  Required  the  square  of  a+b+c+d+e. 

From  this  example  we  infer  that  the  square  of  any  polynomial 
is  composed  of  the  sum  of  the  squares  of  all  the  terms,  together 
with  twice  the  sum  of  the  products  of  all  the  terms  multiplied  to- 
gether two  and  two,  and  this  proposition  may  be  rigorously 
demonstrated. 

It  is  obvious  that  this  rule  for  a  polynomial  includes  the  pre- 
ceding rule  for  a  trinomial,  and  that  in  Art.  60  for  a  binomial. 


SECTION  XL 


EVOLUTION  AND  RADICAL  QUANTITIES. 

(143.)   The  square  root  of  a  quantity  is  a  factor  which,  multi- 
plied by  itself  once,  will  produce  that  quantity. 

Thus,  the  square  root  of  a2  is  a,  because  a  when  multiplied 
by  itself  produces  a2. 

The  square  root  of  144  is  12  for  the  same  reason. 

According  to  Art.  22,  the  square  root  is  indicated  by  the 
sign  V      . 

Thus,  Va*=a, 

and  Vl44a*=12a. 

(144.)  According  to  Art.  137,  in  order  to  square  a  monomial, 
we  must  square  its  coefficient,  and  multiply  the  exponent  of 
each  of  its  letters  by  2.     Therefore,  in  order  to  derive  the 
square  root  of  a  monomial  from  its  square,  we  must 
I.  Extract  the  square  root  of  its  coefficient. 

II.  Divide  each  of  the  exponents  by  2. 

Thus  we  shall  have 


This  is  manifestly  the  true  result,  for 
(8a362)  2  ==  8a862  X  8a362  = 
So,  also, 


For,  (25ab*c*Y=25abtc*X25abica, 

=625aW. 

1.  Required  the  square  root  of  196aa6Vd8. 

2.  Required  the  square  root  of  225a*mb10x*. 

(145.)  According  to  Art.  140,  a  fraction  is  involved  by  in- 
volving both  the  numerator  and  denominator  ;  hence  it  is  ob- 


EVOLUTION    AND    RADICAL    QUANTITIES.  HI 

vious  that  the  square  root  of  a  fraction  is  equal  to  the  root  of 
the  numerator  divided  by  the  root  of  the  denominator. 

Thus  the  square  root  of  rj  is  r. 

4aa 

1.  Find  the  square  root  of-—-:. 

** 


c     a 
2.  rind  the  square  root  of       e  .ia. 

(146.)  It  appears,  from  Art.  144,  that  a  monomial  can  not 
be  a  perfect  square  unless  its  coefficient  be  a  square  number,  and 
the  exponents  of  its  letters  all  even  numbers. 

Thus,  7ab*  is  not  a  perfect  square,  for  7  is  not  a  square  num- 
ber, and  the  exponent  of  a  is  not  an  even  number.  Its  square 
root  may  be  indicated  by  the  usual  sign,  thus,  Vlab*.  Ex- 
pressions of  this  nature  are  called  surds,  or  radicals  of  the  sec- 
ond degree. 

(147.)  We  have  seen,  Art.  137,  that  whatever  may  be  the 
sign  of  a  monomial,  its  square  must  be  affected  with  the  sign 
+.  Hence  we  conclude  that 

If  a  monomial  be  positive,  its  square  root  may  be  either  posi- 
tive or  negative. 

Thus,  V9al=+3a\  or  —  3«3, 

for  either  of  these  quantities,  when  multiplied  by  itself,  pro- 
duces 9a*.  We  therefore  always  affect  the  square  root  of  a 
quantity  with  the  double  sign  ±,  which  is  read  plus  or  minus. 

Thus, 


(148.)  If  a  monomial  be  affected  with  a  negative  sign,  the 
extraction  of  its  square  root  is  impossible,  since  we  have  just 
seen  that  the  square  of  every  quantity,  whether  positive  or 
negative,  is  necessarily  positive. 

Thus,  v/^4,  x/^9,  V-5a, 

are  algebraic  symbols  representing  operations  which  it  is  im- 
possible to  execute.     Quantities  of  this  nature  are  called  im- 
aginary or  impossible  quantities,  and  are  symbols  of  absurdity 
which  we  frequently  meet  with  in  resolving  quadratic  equa 
tions. 

6 


EVOLUTION    AND   RADICAL    QUANTITIES. 

Such  quantities  may  be  represented  by  the  form 
V  —  a,  which  equals 
VaX  —  1=  VaV~l. 

So  tnat  VaV  —  l  is  a  general  form  for  all  imaginary  quan- 
tities of  the  second  degree.     Thus, 

-l=     2   V^l 


That  is,  the  square  root  of  a  negative  quantity  may  always  be 
represented  by  the  square  root  of  a  positive  quantity  multiplied 
by  the  square  root  of  —I. 

(149.)  According  to  Art.  138,  in  order  to  raise  a  monomial 
to  any  power,  we  raise  the  numerical  coefficient  to  the  given 
power,  and  multiply  the  exponent  of  each  of  the  letters  by  the 
exponent  of  the  power  required.  Hence,  reciprocally,  to  ex- 
tract any  root  of  a  monomial,  we  obtain  the  following 

RULE. 

I.  Extract  the  root  of  the  numerical  coefficient. 
II.  Divide  the  exponent  of  each  letter  by  the  index  of  the  re- 
quired root. 

Thus,  ^/64ae&3  =4a*b. 


From  Art.  145,  it  is  obvious  that  to  extract  ANY  root  of  a 
fraction,  we  must  divide  the  root  of  the  numerator  by  the  root 
of  the  denominator. 

„,  ,270663.    3a*b 

Thus  the  cube  root  of  -  is  -  -  ; 
8xsy9      %xy 

Q 

which  may  be  written  -a?bx-*y-\ 

IB 

(150.)  Let  us  now  consider  the  sign  with  which  the  root 
should  be  affected.  We  have  seen,  Art.  139,  that  every  even 
power  is  positive/but  an  odd  power  has  the  same  sign  as  its 
root. 

Thus  —  a,  when  raised  to  different  powers  in  succession, 
will  give 

—  a,  +aa,  —a8,  +a4,  —  <z6,  +«8,  —  a7,  &c.  ; 


EVOLUTION    AND  RADICAL  QUANTITIES.  113 

and  +a,  in  like  manner,  will  give 

+0,  +a\  +a\  +a\  +a>,  +a6,  +a7,&c. 

Since  every  even  number  may  be  expressed  by  2n,  every  even 
power  may  be  considered  as  the  square  of  the  nth  power,  or 
02n=(an)2,  and  must,  therefore,  be  positive  ;  and,  in  like  manner, 
since  an  odd  number  may  be  expressed  by  2n+l,  every  power 
of  an  uneven  degree  may  be  considered  as  the  product  of  the 
2nth  power  by  the  original  quantity,  ancf  must,  therefore,  have 
the  same  sign  with  the  monomial. 

Hence  it  appears, 

I.  An  odd  root  of  any  quantity  must  have  the  same  sign  as  the 
quantity  itself 

Thus, 


II.  An  even  root  of  a  negative  quantity  is  ambiguous. 
Thus,  V8la*bl*=±3ab*. 


III.  An  even  root  of  a  negative  quantity  is  impossible. 

For  no  quantity  can  be  found  which,  when  raised  to  an  even 
power,  can  give  a  negative  result. 

Thus,  \/—  a,  %/—b,  are  symbols  of  operations  which  can 
not  be  performed,  and  they  are  therefore  called  impossible  or 
imaginary  quantities,  as  V  —  «,  in  Art.  148. 

EXAMPLES. 

1.  Find  the  fourth  root  of  8  la8. 

Ans. 

2.  Find  the  fifth  root  of  —243a10&6c-16. 

3.  Find  the  cube  root  of  —  I25aax—'y9. 

4.  Find  the  square  root  of  —  —  . 

5.  Find  the  fifth  root  of 

(151.)  According  to  the  rule  of  Art.  149,  we  perceive  that, 
in  order  that  a  monomial  may  be  a  perfect  power  of  any  degree, 
its  coefficient  must  be  a  perfect  power  of  that  degree,  and  the 

H 


114  EVOLUTION    AND   RADICAL   QUANTITIES. 

exponent  of  each  letter  must  be  divisible  by  the  index  of  the 
root. 

When  the  quantity  whose  root  is  required  is  not  a  perfect 
power  of  the  given  degree,  we  can  only  indicate  the  operation 
to  be  performed.  Thus,  if  it  be  required  to  extract  the  cube 
root  of  4a265,  the  operation  may  be  indicated  by  writing  the 
expression  thus, 


Expressions  of  this  nature  are  called  surds,  or  irrational 
quantities,  or  radicals  of  the  second,  *lhird,  or  nth  degree,  ac- 
cording to  the  index  of  the  root  required. 

(152.)  The  method  of  extracting  the  roots  of  polynomials 
will  be  considered  in  Section  XVII.  There  is,  however,  one 
class  so  simple  and  of  so  frequent  occurrence  that  it  may  prop- 
erly be  introduced  here.  In  Arts.  60  and  61  we  have  seen  that 
the  square  of  a+b  is  a*+2ab-\-b9, 

and  the  square  of       a—  b  is  a2—  2ab-\-b*. 

Therefore,  the  square  root  of  a?±2ab+b*  is  a±b. 

Hence  a  trinomial  is  a  perfect  square  when  two  of  its  terms 
are  squares,  and  the  third  is  the  double  product  of  the  roots  of 
these  squares. 

Whenever,  therefore,  we  meet  with  a  quantity  of  this  de- 
scription, we  may  know  that  its  square  root  is  a  binomial  ;  and 
the  root  may  be  found  by  extracting  the  roots  of  the  two  terms 
which  are  complete  squares,  and  connecting  them  by  the  sign  of 
the  other  term. 

Ex.  1.  Find  the  square  root  of  aa+4a&+46a. 

The  two  terms,  a2  and  4&2  are  complete  squares,  and  the 
third  term  4ab  is  twice  the  product  of  the  roots  a  and  2b  ; 
hence  <z-f  2b  is  the  root  required. 

Ex.  2.  Find  the  square  root  of  9aa—  24ab+16b*. 

Ex.  3.  Find  the  square  root  of  9a*—30a*b+25a'ib\ 

Ex.  4.  Find  the  square  root  of  4a*+I4ab+9b*. 

(153.)  No  binomial  can  be  a  perfect  square.  For  the  square 
of  a  monomial  is  a  monomial  ;  and  the  square  of  a  binomial 
consists  of  three  distinct  terms,  which  do  not  admit  of  being 
reduced  with  each  other. 

Thus  such  an  expression  as 


IRRATIONAL    aUANTITIES.  115 


is  not  a  square  ;  it  wants  the  term  ±2ab  to  render  it  the  square 
of  «±fe.  This  remark  should  be  continually  borne  in  mind 
as  beginners  often  put  the  square  root  of  aa+&a  equal  to  a+b. 


IRRATIONAL  QUANTITIES,  OR  SURDS. 

(154.)  A  rational  quantity  is  one  which  can  be  expressed  in 
finite  terms,  and  without  any  radical  sign  ;  as  a,  5«3,  &c. 

Irrational  quantities,  or  surds,  are  quantities  affected  with  a 
radical  sign,  and  which  have  no  exact  root,  or  a  root  which  can 
be  exactly  expressed  in  numbers. 

Thus,  -v/3  is  a  surd,  because  the  square  root  of  3  can  not  be 
expressed  in  numbers  with  perfect  exactness. 

In  decimals  it  is  1.7320508  nearly. 

(155.)  We  have  seen,  Art.  144,  that  in  order  to  extract  the 
square  root  of  a  monomial,  we  must  divide  each  of  its  expo- 
nents by  2. 

Thus  the  square  root  of  «3  is  a1  or  a ;  that  of  a*  is  a3 ;  that 
of  a6  is  a3,  and  so  on  ;  and  as  this  principle  is  genera],  the  square 

3  5. 

root  of  a3  must  necessarily  be  a~2,  and  that  of  a6  must  be  a2  ; 

and,  in  the  same  manner,  we  shall  have  a*  for  the  square  root 

of  a1.     Whence  we  see  that 

i 
a2  is  equal  to  <ya, 

jj  

cr  is  the  same  as  Va*, 

a?  is  equivalent  to  V an, 

&c.,  &c. 

We  have  also  seen,  Art.  149,  that  in  order  to  extract  any 
root  of  a  monomial,  we  must  divide  the  exponent  of  each  letter 
by  the  index  of  the  required  root. 

Thus,  the  cube  root  of  a8  is  a\  or  a ;  the  cube  root  of  a9  is 
aa ;  the  cube  root  of  a9  is  a3,  and  so  on.  So,  also,  the  cube 

2  4 

root  of  aa  is  a5 ;  the  cube  root  of  a4  is  0s ;  the  cube  root  of  a, 

or  a1,  is  0s.     Whence  it  appears  that 

i_  

a3  is  the  same  as  \/a, 

4  

a3  is  equivalent  to  Va4, 


116  IRRATIONAL    QUANTITIES. 

11  •J^.L 

a?  is  equivalent  to  Va", 

&c.,  &c. 

i 

In  the  same  manner,  the  fourth  root  of  a  is  a¥,  which  expres- 
sion has  therefore  the  same  value  as  Va ;  the  fifth  root  of  a 

will  be  a5,  which  is,  consequently,  equivalent  to  V«,  and  the 
same  principle  may  be  extended  to  all  roots  of  a  higher  de- 
gree. 

(156.)  Other  fractional  exponents  are  to  be  understood  in 

5 

the  same  way.  Thus,  if  we  have  a1,  this  means  that  we  must 
first  take  the  fifth  power  of  a,  and  then  extract  its  fourth  root ; 

so  that  a?  is  the  same  as  Va6. 

m 

So,  also,  to  find  the  value  of  a",  we  must  first  take  the  mth 
power  of  a,  which  is  am,  and  then  extract  the  nth  root  of  that 

m 

power ;  so  that  a"  is  the  same  as  Vam. 

Hence  the  numerator  of  a  fractional  exponent  denotes  the 
power,  and  the  denominator  the  root  to  be  extracted. 

Again,  let  it  be  required  to  extract  the  cube  root  of  — . 

In  the  first  place,  —  =a~\     Now,  to  extract  the  cube  root 
of  a~*,  we  must  divide  its  exponent  by  3,  which  gives  us 


But  the  cube  root  of —may  also  be  represented  by  — . 

1  _4 

Hence  ~r  is  equivalent  to  a  ?. 
a? 

So,  also,  ~i  is  equivalent  to  a~"% 
a2 

~~l  is  equivalent  to  a~", 
an 

J_  -™ 

~»  is  equivalent  to  a  n. 
a" 

Thus  we  see  that  the  principle  of  Art.  69,  that  a  factor  maj 
be  transferred  from  the  numerator  to  the  denominator  of  a 


IRRATIONAL    QUANTITIES.  117 

fraction,  or  from  the  denominator  to  the  numerator  by  chang 
ing  the  sign  of  its  exponent,  is  applicable  also  to  fractional  ex 
ponents. 

We  may  therefore  entirely  reject  the  radical  signs  hitherto 
made  use  of,  and  employ,  in  their  stead,  the  fractional  expo- 
nents which  we  have  just  explained  ;  and,  indeed,  many  of  the 
difficulties  in  the  reduction  of  radical  quantities  disappear  when 
fractional  exponents  are  substituted  for  the  radical  signs. 


PROBLEM  I. 

To  reduce  surds  to  their  most  simple  forms. 

(157.)  Surds  may  frequently  be  simplified  by  the  application 
of  the  following  principle  :  the  square  root  of  the  product  of  two 
or  more  factors  is  equal  to  the  product  of  the  square  roots  of 
those  factors. 

Or,  in  algebraic  language, 

Vab=  v/aX  Vb. 

For  each  member  of  this  equation  squared  will  give  the 
same  quantity. 

Thus,  the  square  of  Vab  is  ab. 

And  the  square  of  «/aX  Vb  is  (VaYx(Vby=ab. 

Hence,  since  the  squares  of  the  quantities  Vab  and  */aX  \/b 
are  equal,  the  quantities  themselves  must  be  equal. 

Let  it  be  required  to  reduce  V4a  to  its  most  simple  form. 

This  expression  may  be  put  under  the  form  x/4X  %/«. 

But  x/4  is  equal  to  2. 

_  j_ 

Hence,  V4a=^/4X  ^/a=2^/a=2a2. 

2x/«  is  considered  a  simpler  form  than  V4a,  for  reasons  whicn 
will  be  better  understood  hereafter. 

Again,  reduce  </48  to  its  most  simple  form. 
x/48  is  equal  to 


Therefore,  in  order  to  simplify  a  monomial  radical  of  the 
second  degree,  separate  it  into  two  factors,  one  of  which  is  a 
perfect  square;  extract  its  root;  and  prefix  it  to  the  other  factor 
with  the  radical  sign  between  them. 

In  the  expressions  2  ^/a  and  4v/3,  the  quantities  2  and  4  are 
called  the  coefficients  of  the  radical. 


118  IRRATIONAL    aUANTITIES. 

EXAMPLES. 

1.  Reduce  2</32  to  its  most  simple  form 


2.  Reduce  Vl25a*  to  its  most  simple  form. 

Ans. 

3.  Reduce  V98ab*  to  its  most  simple  form. 

Ans.  W  V2a. 

4.  Reduce  V294«&a  to  its  most  simple  form. 

5.  Reduce  7V80a6c8  to  its  most  simple  form. 

6.  Reduce  V98a*x'y*  to  its  most  simple  form. 

7.  Reduce  V45aa&3cad  to  its  most  simple  form. 

8.  Reduce  V864a865cn  to  its  most  simple  form. 

(158.)  Surds  of  any  degree  may  be  simplified  by  the  appli- 
cation of  the  following  principle,  which  is  merely  an  extension 
of  that  already  proved  in  the  preceding  Article. 

The  nth  root  of  the  product  of  any  number  of  factors  is  equal 
to  the  product  of  the  nth  roots  of  those  factors. 

Or,  in  algebraic  language, 

V~ab=VaX  Vb. 

For,  raise  each  of  these  expressions  to  the  nth  power,  and  we 
shall  obtain  the  same  result. 

Thus,  the  nth  power  of  Vdb  is  ab. 

And  the  rath  power  of  VaX  Vb  is  (Va)"X(Vb)n=ab. 

Hence,  since  the  same  powers  of  the  quantities  Vab  and 
VaX  Vb  are  equal,  the  quantities  themselves  must  be  equal. 

Let  it  be  required  to  reduce  V8aa  to  its  most  simple  form. 

This  is  equivalent  to  V8X  Va2,  which  is  equal  to  2  Vaa. 

Again,  take  the  expression 


This  is  equivalent  to  Vl6a4X  V3a,  which  is  equal  to  2a  V~3a. 
Hence,  to  simplify  a  monomial  radical  of  any  degree,  we 
have  the  following 

RULE. 
Separate  the  quantity  into  two  factors,  one  of  which  is  an  ex- 


IRRATIONAL    aUANTITIES.  119 

act  power  of  the  same  name  with  the  root ;  extract  its  root ;  and 
prefix  it  to  the  other  factor  with  the  radical  sign  between  them. 

In  the  expressions  2Va2  and  2a  V3a,  the  quantities  2  and  2a 
placed  before  the  radical  sign  are  called  the  coefficients  of  the 
radical. 

EXAMPLES. 


1.  Reduce  \/56«566  to  its  most  simple  form. 

Ans. 

2.  Reduce  \/54a46V  to  its  most  simple  form. 

Ans. 


3.  Reduce  V48a&6V  to  its  most  simple  form. 

Ans.  2ab*cV~3ac\ 


4.  Reduce  Vl92«7&cia  to  its  most  simple  form. 

5.  Reduce  Vl92«46V  to  its  most  simple  form. 

6.  Reduce  9  V81&2  to  its  most  simple  form. 

(159.)  There  is  another  principle  which  can  frequently  be 
employed  to  advantage  in  simplifying  radicals. 

The  square  of  the  cube  of  a  is  equal  to  the  sixth  power  of  a. 

For  the  square  of  the  cube  of  a  is  a3  X  a3, 
which  equals  a3+3=a6. 

So,  also,  the  fourth  power  of  the  cube  of  a  is  equal  to  the 
twelfth  power  of  a. 

For  («3)4=a3Xa3Xa3Xa3 


And,  in  general,  the  mth  power  of  the  nth  power  of  any 
quantity  is  equal  to  the  mnth  power  of  that  quantity. 
That  is  (an)m=ann. 
Hence,  conversely, 

The  mnth  root  of  any  quantity  is  equal  to  the  mth  root  oi 
the  nth  root  of  that  quantity. 

Thus,  the  fourth  root  =  the  square  root  of  the  square  root  ; 
"      the  sixth  root  =  the  square  root  of  the  cube  root,  or 

the  cube  root  of  the  square  root  ; 
u      the  eighth  root  =  the  square  root  of  the  fourth  root,  or 

the  fourth  root  of  the  square  root  ; 
•*      the  ninth  root  =  the  cube  root  of  the  cube  root. 
6* 


120  IRRATIONAL    dUANTITIES. 

Hence,  when  the  index  of  a  root  is  the  product  of  two  or  more 
factors,  we  may  obtain  the  root  required  by  extracting  in  suc- 
cession the  roots  denoted  by  those  factors. 

Ex.  1.  Let  it  be  required  to  extract  the  sixth  root  of  64. 

The  sixth  root  is  equal  to  the  cube  root  of  the  square  root. 

The  square  root  of  64  is  8, 
and  the  cube  root  of  8  is  2. 

Hence  the  sixth  root  of  64  is  2. 

Ex.  2.  Let  it  be  required  to  extract  the  eighth  root  of  256. 

The  eighth  root  is  equal  to  the  fourth  root  of  the  square  root ; 
or  to  the  square  root  of  the  square  root  of  the  square  root. 

The  square  root  of  256  is  16, 
and  the  fourth  root  of  16  is  2. 

Hence  the  eighth  root  of  256  is  2. 

When  one  of  the  roots  can  be  extracted,  and  the  other  can 
not,  a  radical  may  be  simplified  by  extracting  one  of  the  roots. 

Thus,  the  fourth  root  of  9  is  equal  to  the  square  root  of  the 
square  root  of  9  ;  that  is,  the  square  root  of  3. 

Or,  algebraically,  V9=  V3. 

Ex.  3.  Reduce  V4aa  to  its  most  simple  form. 

Ans.  %/2a. 

Ex.  4.  Reduce  V36a362  to  its  most  simple  form. 

Ex.  5.  Reduce  mtyan  to  its  most  simple  form. 

Ex.  6.  Reduce  V25a46V  to  its  most  simple  form. 

PROBLEM  II. 

(160.)   To  reduce  a  rational  quantity  to  the  form  of  a  surd. 
The  square  root  of  the  square  of  a  is  obviously  a ;  that  is, 

a=  Vtf=a*. 
So,  also,  the  cube  root  of  the  cube  of  a  is  a  ; 

that  is,  a=  Va'=a* 

Hence,  to  reduce  a  rational  quantity  to  the  form  of  a  surd, 
we  have  the  following 

RULE. 

Raise  the  quantity  to  a  power  of  the  same  name  with  the  given 
rooty  and  then  apply  the  corresponding  radical  sign. 


IRRATIONAL    aUANTITlES.  121 

EXAMPLES. 

1.  Reduce  3  to  the  form  of  the  square  root. 

Here  3X3=32=9;  whence  3=  v/9.  Ans. 

2.  Reduce  ax  to  the  form  of  the  square  root. 


Ans.   VoV,  or  (aV)*. 

3.  Reduce  2x*  to  the  form  of  the  cube  root. 

Ans.   y~8x6. 

4.  Reduce  5+6  to  the  form  of  the  square  root. 

5.  Reduce  —  3x  to  the  form  of  the  cube  root. 

6.  Reduce  —  ±xz  to  the  form  of  the  fourth  root. 

7.  Reduce  asb*  to  the  form  of  the  square  root. 

8.  Reduce  am  to  the  form  of  the  nth  root. 

It  will  be  observed,  that  this  Problem  is  nearly  the  reverse  of 
the  preceding,  and,  consequently,  brings  quantities  into  a  less 
simple  form  ;  nevertheless,  this  form  is  sometimes  better  suited 
to  subsequent  operations,  as  will  be  seen  hereafter. 


PROBLEM  III. 

(161.)  To  reduce  surds  which  have  different  indices  to  others 
of  the  same  value  having  a  common  index. 

Ex.  1.  Reduce  a2  and  a3  to  surds  having  the  same  radical 
sign. 

From  the  preceding  Article,  it  is  obvious  that  the  square 
root  of  a  is  equal  to  the  sixth  root  of  the  cube  of  a  ; 

that  is,  a?=a?=  ty~a*. 

So,  also,  a*=a*=  vV. 

Thus,  the  quantities  a2  and  aj  are  reduced  to  \/«3  and  V«", 
which  are  of  the  same  value,  and  have  the  common  index  6. 

Ex.  2.  Reduce  32  and  2s  to  a  common  index. 


Hence  V27  and  -^4  are  the  quantities  required. 
Whence  we  derive  the  following 


122  IRRATIONAL    QUANTITIES. 

RULE. 

Reduce  the  fractional  exponents  to  a  common  denominator ; 
raise  each  quantity  to  the  power  denoted  by  the  numerator  of  its 
reduced  exponent ;  and  take  the  root  denoted  by  the  common  de- 
nominator. 

Ex.  3.  Reduce  27  and  4T  to  a  common  index. 

Ans.    V4  and  V8. 

Ex.  4.  Reduce  aa  and  a*  to  a  common  index. 

±  ^ 

Ex.  5.  Reduce  a2  and  b3  to  a  common  index. 

2  3 

Ex.  6.  Reduce  5s  and  7¥  to  a  common  index. 
Ex.  7.  Reduce  a"  and  bm  to  a  common  index. 

PROBLEM  IV. 

To  add  surd  quantities  together. 

(162.)  Two  radicals  are  similar  when  they  have  the  same 
index,  and  the  same  quantity  under  the  radical  sign. 

Thus,  3  Va  and  5  Va  are  similar  radicals. 

So,  also,  7V&  and  10  Vb  are  similar  radicals. 

But  Va  and  \fa  are  not  similar  radicals  ;  for,  although  they 
have  the  same  quantity  under  the  radical  sign,  they  have  not 
the  same  index. 

Ex.  1.  Find  the  sum  of  2V a  and  3V a. 

As  these  are  similar  radicals,  we  may  unite  their  coefficients 
by  the  usual  rule  ;  for  it  is  evident  that  twice  the  square  root  of 
a  and  three  times  the  square  root  of  a  make  five  times  the 
square  root  of  a.  Hence  the  following 

RULE. 

When  the  radicals  are  similar,  add  the  coefficients,  and  annex 
the  radical  part. 

But  if  the  quantities  are  dissimilar,  and  can  not  be  made  sim- 
ilar by  the  reductions  in  the  preceding  Articles,  they  can  only  be 
connected  together  by  the  sign  of  addition. 

Ex.  2.  Add  V6  to  2V 6. 

Ans.  3V  6. 


IRRATIONAL    QUANTITIES.  123 

Ex.  3.  Add  5Va  and  -2V a. 


Ex.  4.  Add  aVb+c  and  xVb+c. 

If  the  radical  parts  are  originally  different,  they  must,  if  pos- 
sible, be  made  alike  by  the  preceding  methods. 

Ex.  5.  Add  V27  to  V48. 

Here  \/27=  V  9X3=3  V3, 

and 


Whence  their  sum  =7\/3. 

j&c.  6.  Add  together  V500  and  V108. 

Ans.  8V4. 
Ex.  7.  Add  together  4V  147  and  3V75. 

Ans.  43  VS. 

Ex.  S.  Add  together  3  Vf  and  2V~^. 
Here  3^/f  =3^«  =j   v/10, 

and 


Whence  their  sum  =f  x/10. 

JKa;.  9.  Add  together  v/72  and  x/128. 
Ea;.  10.  Add  together  v/180  and  ^405. 
jE«.  11.  Add  together  V40  and  V135. 
Ex.  12.  Add  together  8V32  and  5V2. 

PROBLEM  V. 

To  find  the  difference  of  surd  quantities. 

(163.)  It  is  evident  that  the  subtraction  of  surd  quantities 
may  be  performed  in  the  same  manner  as  addition,  except  that 
the  signs  in  the  subtrahend  are  to  be  changed  according  to 
Art.  43. 

Ex.  1.  Required  to  find  the  difference  between  </448  and 
v/112.  _ 

Here  ^448=  V64X  7=8^7, 


and  ^112=  Vl6X7=4v/7. 

Whence  the  difference 


Ex.  2.  Find  the  difference  between  V192  and  V24. 
Here  V  192=  V64~x¥=4  V3, 


and  V24  =  V  8x3=2^3. 

Whence  the  difference  =2V3. 


124  IRRATIONAL    QUANTITIES. 

Ex.  3.  Find  the  difference  between  5v/20  and  3^45. 

Here  5  </20=5\/4X5=  10  V5, 

and  3V45=3V9X5=  9V  5. 

Whence  the  difference  =  V5. 

Ex.  4.  Find  the  difference  between  2^50  and  -v/18. 

Ex.  5.  Find  the  difference  between  2^320  and  3^40. 


Ex.  6.  Find  the  difference  between  V8Qa*x  and  V20aV. 


Ex.  7.  Find  the  difference  between  2\/72aa  and 


PROBLEM  VI. 

To  multiply  surd  quantities  together. 

(164.)  Let  it  be  required  to  multiply  Va  by  Vb. 

The  product  will  be  Vab. 

For  if  we  raise  each  of  these  quantities  to  the  power  of  n, 
we  obtain  the  same  result,  ab  ;  hence  these  two  expressions 
are  equal.  We  therefore  have  the  following 


RULE. 


.  -  • 

When  the  surds  have  the  same  index,  multiply  the  quantities 
under  the  sign  by  each  other,  and  prefix  the  common  radical 
sign.  If  there  are  coefficients,  these  must  be  multiplied  separ- 
ately. 

Ex.  1.  Required  the  product  of  3  V  8  and  2VQ. 
Here  3V  8, 

multiplied  by  2V  6,        __ 

gives  GV4S=GVlQXB=24V3.tAns. 

Proof.  Square  3</8,  and  we  obtain  9X8=72. 
Square  2v/6,  and  we  obtain  4X6=24. 
72  multiplied  by  24=1728. 
Also,  24^3  squared  =576X3=1728. 
Ex.  2.  Required  the  product  of  5  V8  and  3  V5. 

Ans.  30/10. 
Ex.  3.  Required  the  product  of  7^18  and  5^4. 

Ans. 

Ex.  4.  Required  the  product  of  1^6  and  f^n. 
Ex.  5.  Required  the  product  of  ?,  v'lS  and  5^20. 


IRRATIONAL    Q.UANTITIES.  125 

In  the  preceding  examples,  let  all  the  results  be  reduced  to 
their  simplest  form. 

If  the  surds  have  not  the  same  index,  they  must  first  be  re- 
duced to  a  common  index,  by  Art.  161. 

Ex.  6.  Required  the  product  of  </2  and  #3. 

Here  ^2=2*=(2«)*=  V  8, 

and  ^3=3f=(32)*=V  9. 

Whence  the  product  ^72. 

(165.)  We  have  seen,  in  Art.  50,  that  powers  of  the  same 
quantity  may  be  multiplied  by  adding  their  exponents.  The 
same  principle  may  be  extended  to  roots  of  the  same  quantity. 

Let  it  be  required  to  multiply  Va  by  V«,  or  a2  by  a3. 

J.  3  12 

We  have  seen,  in  Art.  161,  that  a2=a*,  and  a*=a*. 

3111 

But  a*=avXa?Xav, 
and      a?=a*Xals. 

111115 

The  product,  therefore,  is  asXa*XasXa*Xa*=ae. 
Hence,  roots  of  the  same  quantity  may  be  multiplied  by  add- 
ing their  fractional  exponents. 

Ex.  1.  Multiply  5a¥  by  So5. 

Ans.  I5a\ 

Ex.  2.  Multiply  So*  by  21a*.  " 
Ex.  3.  Multiply  3x*y*  by 


Ex.  4.  Multiply  (a+b)"  by  (a+b)". 

(166.)  If  the  rational  quantities,  instead  of  being  coefficients 
of  the  radical  quantities,  are  connected  with  them  by  the  signs 
+  or  —  ,  each  term  of  the  multiplier  must  be  multiplied  into 
each  term  of  the  multiplicand. 

1.  Let  it  be  required  to  multiply  8+    </5 
by  2-    ^5 

6+2^5 

-3y/5-5. 

We  obtain  the  product  6—    v/5—  5, 

which  reduces  to  1—   v/5. 


126  IRRATIONAL    QUANTITIES. 

2.  Multiply  7+2^6  by  9-5^6. 

Ans.  3-17v/6. 

3.  Multiply  9+2v/10  by  9-2v/10. 

Ans.  41. 


PROBLEM  VII. 

To  divide  one  surd  quantity  by  another. 

(167.)  Let  it  be  required  to  divide  tya*  by  Va\ 

The  quotient  must  be  a  quantity  which,  multiplied  by  the 

divisor,  shall  produce  the  dividend;  we  thus  obtain  Va;  for, 

according  to  Art.  164,  Va'X  Va--  Va*  ; 

that  is, 

Hence  the  following 

RULE. 

Quantities  under  the  same  radical  sign  may  be  divided  like 
rational  quantities,  the  quotient  being  affected  with  the  common 
radical  sign.  If  there  are  coefficients,  they  must  be  divided 
separately. 

If  the  radicals  have  not  the  same  index,  we  must  first  reduce 
them  to  a  common  index. 

• 

EXAMPLES. 

1.  It  is  required  to  divide  8  -/1  08  by 


Here  =4  ^  18=4  V9^<2=  12^2.  Ans. 

2.  Divide  8V512  by  4V2. 

Here        ^;p=2  V256=2V64X4=8  V4.  Ans. 

3.  Divide  6  V54  by  3^2. 

Ans.  6. 

4.  Divide  4  V72  by  2V18. 

5.  Divide  4  V  ~Wy  by  2  V~3y. 

Ans.  2aV2. 

L  L 

6.  Divide  16(a'6)-  by  8(ac)"*  . 

7.  Divide  4V  12  by 


IRRATIONAL    QUANTITIES.  127 

As  the  radicals  in  this  last  example  have  not  the  same  in- 
dex, they  must  be  reduced  to  a  common  index. 

4Vl2=4(12f=4(12)l=4(144)l 
2v/3  =2(3)*  =2(3)*  =2(27)1 

4(1441*  l  l 

Hence         -*  -  }-=2(\^  =2(^=2  Vy. 

2(27)¥ 

(168.)  We  have  seen,  in  Art.  67,  that,  in  order  to  divide 
quantities  expressed  by  the  same  letter,  we  must  subtract  the 
exponent  of  the  divisor  from  the  exponent  of  the  dividend. 
The  same  principle  may  be  extended  to  fractional  exponents. 

Thus,  let  it  be  required  to  divide  a?  by  a*. 
According  to  the  preceding  Article, 

* 


Hence  a  root  is  divided  by  another  root  of  the  same  letter  01 
quantity,  by  subtracting  the  exponent  of  the  divisor  from  that 
of  the  dividend. 

-Ex.l.  Divide  (abf  by  (06)* 

Ans. 
Ex.  2.  Divide  a3  by  a*. 

Ex.  3.  Divide  a*  by  a*. 
i          i 
Ex.  4.  Divide  a"  by  am. 

Ex.  5.  Divide  4^/ab  by  2  Vab. 

Ans. 

PROBLEM  VIII. 

(169.)  To  raise  surd  quantities  to  any  power. 

_i 

Let  it  be  required  to  find  the  square  of  a?. 
The  square  of  a  quantity  is  found  by  multiplying  it  by  itself 
once. 

1  1  1  Li.1  2 

Hence  the  square  of  a3  is  equal  to  a7Xa^=a    Tr=a7. 

(1\  3  2 

a3)  =a3. 


128  IRRATIONAL    dUANTITIES. 

i 

Again,   let  it  be  required  to  find  the  cube  of  a5. 
The  cube  of  a  quantity  is  found  by  multiplying  it  by  itself 
twice. 

Hence  the  cube  of  a5  is  equal  to  asXasXas=as  ; 

(1\  3  3 

a5)  =a5. 

L       " 
In  the  same  manner  we  should  find  the  nth  power  of  am=an. 

Hence  we  have  the  following 

RULE. 

Radical  quantities  are  involved  by  multiplying  their  fractional 
exponents  by  the  exponent  of  the  required  power. 

Ex.  1.  Required  the  fourth  power  of  fa3. 
Ex.  2.  Required  the  cube  of  f  \/3. 

Ans.  fv/3. 
Ex.  3.  Required  the  square  of  3  1/3. 

Ex.  4.  Required  the  cube  of  17-/21. 

Ex.  5.  Required  the  fourth  power  of  |\/6. 

Ans.  3^' 

(170.)  If  the  radical  quantities  are  connected  with  others  by 
the  signs  -\-and  —  ,  they  must  be  involved  by  a  multiplication  of 
the  several  terms. 

Ex.  1.  Required  the  square  of  3+  v/5. 

3+   x/5 


The  square  is  9+6^/5+5 

or  14+6^5.  Ans. 

Ex.  2.  Required  the  square  of  3+2^/5. 

These  two  examples  are  comprehended  under  the  rule  in 
Art.  60,  that  the  square  of  the  sum  of  two  quantities  is  equal 
to  the  square  of  the  first,  plus  twice  the  product  of  the  first  by 
the  second,  plus  the  square  of  the  second. 

Ex.  3.  Required  the  cube  of  </x+3^/y. 

Ex.  4.  Required  the  fourth  power  of  v/3—  v/2. 

Ans.  49-20V6. 


IRRATIONAL    QUANTITIES.  129 

PROBLEM  IX. 

To  find  the  roots  of  surd  quantities. 

(171.)  A  root  of  a  quantity  is  a  factor  which,  multiplied  by 
itself  a  certain  number  of  times,  will  produce  the  given  quan- 
tity. But  we  have  seen  that  a  radical  quantity  is  involved  by 
multiplying  its  exponent  by  the  exponent  of  the  required  pow- 
er. Hence, 

To  find  the  roots  of  surd  quantities, 

Divide  the  fractional  exponent  by  the  index  of  the  required 
root. 

Thus,  the  square  root  of  a?  is  as~r*=a*. 

For,  by  Art.  169,  we  obtain  the  square  of  a*  by  multiplying 
the  exponent  F  by  2  ; 
that  is,  (a*y=a*=a\ 

EXAMPLES. 

1.  Find  the  square  root  of  9(3)^. 

Here         (9(8)*)*=9JX8*4*=8(8)*=SV8.  Ans. 

2.  Required  the  cube  root  of  jv/2. 

Ans.  £V2. 

3.  Required  the  square  root  of  103. 

4.  Required  the  cube  root  of  /T#4. 

ij 

5.  Required  the  fourth  root  o 

6.  Required  the  cube  root  o 

7.  Required  the  cube  root  of  fv/f. 

Ans.   y/l- 

PROBLEM  X. 

To  find  multipliers  which  shall  cause  surds  to  become  rational. 

(172.)  I.  When  the  surd  is  a  monomial. 

The  quantity  ^a  is  rendered  rational  by  multiplying  it  by 


For 


130  IRRATIONAL    QUANTITIES. 

So,  also,  0s  is  rendered  rational  by  multiplying  it  by  a3. 
For  a?Xc?=c?=a. 

I  3 

Also,       of  is  rendered  rational  by  multiplying  it  by  a*. 
For  afXa(=af=a. 

I  n— I 

In  general,  <r  is  rendered  rational  by  multiplying  it  by  a  n  . 

1  n—l  n—l+l  n 

For  anXa  n  —a  n    =a"—a. 

Hence  we  deduce  the  following 

RULE. 

Multiply  the  surd  by  the  same  quantity  having  such  an  ex 
ponent  as,  when  added  to  the  exponent  of  the  given  surd,  shall  bt 
equal  to  unity. 

(173.)  II.  When  the  surd  is  a  binomial 

If  the  binomial  contains  only  the  square  root,  multiply  the 
given  binomial  by  the  same  expression  with  the  sign  of  one  of  its 
terms  changed,  and  it  will  give  a  rational  product. 

Ex.  1.  The  expression  */a+^b 

Multiplied  by  v/a—  ^/b 

a-\-  Vab 

-Vab-b 
Gives  a  product  a.  —b,  which  is  rational. 

Ex.  2.  Find  a  multiplier  which  shall  render  5+  v/3  rational 

Given  surd,  5+  x/3 

Multiplier,  5—^3 

Product,  25—3=22,  as  required. 

These  two  examples  are  comprehended  under  the  Rule  in 
Art.  62,  the  product  of  the  sum  and  difference  of  two  quantities 
is  equal  to  the  difference  of  their  squares. 

Ex.  3.  Find  a  multiplier  that  shall  make  -v/5+^3  rational, 
and  determine  the  product. 

Ex.  4.  Find  a  multiplier  that  shall  make  \/5—  ^/x  rational, 
and  determine  the  product. 

Ex.  5.  Find  a  multiplier  that  shall  make  Va—  Vabc  ra- 
tional. 


IRRATIONAL    QUANTITIES.  131 

ill.  When  the  surd  is  a  trinomial,  it  may  be  reduced,  by 
successive  multiplications,  first  to  a  binomial  surd,  and  then  to 
a  rational  quantity. 

Ex.  1.  Find  multipliers  that  shall  make  v/5-f  <v/3—  >/2  ra- 
tional. 

Given  surd,  v/5+  v/3  —  v/2 

First  multiplier,      v/5+x/3+v/2 


3—  x/6 
+N/10+X/6-2 


_ 

First  product,  2  x/  1  5  +  6 

Second  multiplier,         2  -/1  5—  6 

60+12x/15 

-12V/15-36 

Second  product,  60—36=24,  a  rational  quantity. 

£!#.  2.  Find  multipliers  that  shall  make  ^/a+^/b+^/c  ra- 
tional, and  determine  the  product. 


PROBLEM  XL 

(174.)   To  reduce  a  fraction  containing  surds  to  another  hav- 
ing a  rational  numerator  or  denominator. 

RULE. 

Multiply  both  numerator  and  denominator  by  a  factor  which 
will  render  either  of  them  rational,  as  the  case  may  require. 

Ex.  1.  If  both  terms  of  the  fraction  —  -  be  multiplied  by 
•v/a,  it  will  become  -  r,  in  which  the  numerator  is  rational. 


Or  if  both  terms  be  multiplied  by  -/&,  it  will  become  —  r—  ,  in 

which  the  denominator  is  rational. 

2 

Ex.  2.  Reduce  the  fraction  —  —  to  one  that  shall  have  a  ra- 

v*> 

tional  denominator. 

2v/3 
Ans.  —  . 


132  IRRATIONAL    ClUANTITIES. 

Ex.  3.  Reduce  —  -  -  -  to  a  fraction  having  a  rational  de- 
\/5  —  </£ 

nominator. 


/2 
Ex.  4.  Reduce  -  -  -  to  a  fraction  having  a  rational  de- 

o  —  v^ 

nominator. 


Ex.  5.  Reduce  ,  to  a  fraction  having  a  rational  de- 

nominator. 

4 
JBx.  6.   Reduce  —  ^-:  —  to   an   expression   having   a 

V  "i    V  -w"!"! 

rational  denominator. 

Ans.  2+V/2—  v/6. 

JEfo.  7.  Reduce  x/5-f-\/2  to  a  fraction  having  a  rational 
numerator. 

(175.)  The  utility  of  the  preceding  transformations  may  be 
illustrated  by  computing  the  numerical  value  of  a  fractional 
surd. 

Ex.  1.  Suppose  it  is  required  to  find  the  square  root  of  ^  ; 

/o 

that  is,  it  is  required  to  find  the  value  of  the  fraction  —  . 

v  ' 


If  we  make  the  denominator  rational,  we  shall  have  —  —  ,  in 

which  it  is  only  necessary  to  extract  the  square  root  of  the 
numerator,  and  the  value  of  the  fraction  is  found  to  be  0.6546. 

*7     /  1 

Ex.  2.  It  is  required  to  find  the  value  of  the  fraction  —j  -  . 

Making  the  denominator  rational,  we  have  -  —    —,  the 

value  of  which  is  3.1003. 

</6 
Ex.  3.  Required  the  value  of  the  expression  -—=-;  —  —. 

V  '  ~T~  v  " 
Ans.  0.5595. 


IRRATIONAL    UUANTITIES.  133 

Ex.  4.  Required  the  value  of  the  expression 


2v/8+3v/5-7v/2* 

Ans.  0.7025. 


^ 
Ex.  5.  Required  the  value  of  the  expression  9_^      Q. 

.  5.7278. 


PROBLEM  Xli. 

(176.)   To  free  an  equation  from  radical  quantities. 

This  may  generally  be  done  by  successive  involutions.  For 
this  purpose,  we  first  free  the  equation  from  fractions.  If  there 
is  but  one  radical  expression,  we  bring  that  to  stand  alone  on 
one  side  of  the  equation,  and  involve  the  whole  equation  to  a 
power  denoted  by  the  index  of  the  radical. 

Ex.  1.  Free  the  equation 


a 

from  radical  quantities. 
Clearing  of  fractions,  and  transposing  a,  we  obtain 

V2ax+x*=ab—  a. 
The  square  of  this  equation  is 


which  is  free  from  radical  quantities. 
Ex.  2.  Free  the  equation 

_  20' 


from  radical  quantities. 

If  the  equation  contains  two  radical  expressions,  combined 
with  other  terms  which  are  rational,  it  will  generally  be  best 
to  bring  one  of  the  radicals  to  stand  alone  on  one  side  of  the 
equation  before  involution.  One  of  the  radicals  will  thus  be 
made  to  disappear,  and,  by  repeating  the  operation,  the  re- 
maining radical  may  be  exterminated. 

Ex.  3.  Free  the  equation 

Va+x-t-  Vb+y=c 
from  radical  quantities. 


134  IRRATIONAL    aUANTITIES. 

Transposing  one  of  the  radicals,  we  obtain 

Va+x=c—  Vb+y. 
Squaring,  we  have 

a+x=c*—2c  Vb+y+b+y. 

Transposing,  so  as  to  bring  the  radical  to  stand  alone,  we 
have 

2c  Vb+y=c*+b+y-a—x, 
which  may  be  freed  from  radicals  by  squaring  a  second  time. 

Sometimes  the  two  radicals  may  be  of  such  a  form  that  il 
is  best  to  bring  both  to  the  same  member  of  the  equation  be- 
fore involution. 

When  an  equation  contains  several  radical  quantities,  it  may 
generally  be  freed  from  them  by  successive  involutions,  bul 
the  best  mode  of  procedure  can  only  be  determined  by  trial. 

Ex.  4.  Free  the  equation 


V3x-lS=  Vlx+I 
from  radical  quantities. 

Ans.  6x2—  I5x—  12G=x*-t-12x+36. 

When  an  equation  contains   a  fraction   involving  radical 
quantities  in  both  numerator  and  denominator,  it  is  sometimes 
best  to  render  the  denominator  rational  by  Problem  XL 
Ex.  5.  Free  the  equation 

Vx-\-Vx—a      an* 


Vx—  V.x—a    x—& 
from  radical  quantities. 

Multiply  both  terms  of  the  first  fraction  by  Vx+  Vx—a,  and 
we  have 

(Vx+  Vx— a)*_  an* 
x—(x—a)        x—a9 

03^3 

or  (Vx+  \fx— «)a= . 

X~~~CL 

Extracting  the  square  root,  we  obtain 

an 

Vx+Vx—a= 


Vx—a 
Clearing  of  fractions,  we  have 

Vx9 — ax +x — a = an, 
which  is  easily  freed  from  radicals. 


IRRATIONAL    QUANTITIES.  135 

Ex.  6.  Free  the  equation 

x+  v/#_o;a—  x 
x—  Jx~     4 
from  radical  quantities. 

Ans. 
Ex.  7.  Free  the  equation 

x— 

x+Vx+l 
from  radical  quantities. 

Ans.  9#a= 
Ex.  8.  Free  the  equation 


from  radical  quantities. 

(177.)  The  preceding  rules  for  the  reduction  of  radicals,  are 
exact  so  long  as  we  treat  of  absolute  numbers,  but  require 
some  modifications  when  we  consider  imaginary  expressions, 
such  as  V  —  3,  V—  a,  &c. 

Let  it  be  required,  for  example,  to  determine  the  product  of 
V  —  a  by  V—  a. 

By  the  rule  given  in  Art.  164, 


V—aX  V  —  a=  V—aX—a 


Now,  V  +a*=±a,  so  that  there  is  apparently  a  doubt  as  to 
the  sign  with  which  a  ought  to  be  affected  in  order  to  answer 
the  question.  However,  the  true  result  is  —  «,  because  any 
quantity  must  be  equal  to  the  square  of  its  square  root. 

That  is,  V—axV  —a  is  the  same  as  (V—  «)2,  and,  conse- 
quently, is  equal  to  —  a. 

Again,  let  it  be  required  to  determine  the  product  of  V—  a 
by  V^-b. 

By  the  rule  in  Art.  164. 


V—  ax  Vb=  V  —  aX—b 


The  result,  however,  is  not  properly  ambiguous,  and  should 
be  —  Vab  ;  for  we  have,  according  to  Art.  148, 

7 


136  IRRATIONAL    ClUANTITIES. 

V  —  a=  Va.  y/  —  1, 

and  V^b=Vb.V^l. 

Hence 


=  VabX  — 


In  the  same  manner  we  shall  find  for  the  different  powers 
of  V^l  the  following  results. 

V^l    =  V  —  1,  the  first  power. 
(V  —  l)a=  —  1,  the  second  power. 

(V~^iy=-ixV^i 

=  —</  —  !,  the  third  power. 


=  +  1,  the  fourth  power. 

Since  the  four  following  powers  will  be  found  by  multiply- 
ing —  1  by  the  first,  the  second,  the  third,  ^and  the  fourth 
powers,  we  shall  again  find  for  the  four  next  powers 

+  V^I,  -1,  -V^T,  +1; 

so  that  all  the  powers  of  V  —  1  will  form  a  repeating  cycle  of 
these  four  terms. 

Whenever  the  student  is  at  a  loss  to  determine  the  product 
of  two  imaginary  quantities,  it  is  best  to  resolve  each  of  them 
into  two  factors,  one  the  square  root  of  a  positive  quantity,  and 
the  other  V^l,  Art.  148. 

EXAMPLES. 

1.  Let  it  be  required  to  multiply  V—  9  by  V—  4. 
Here  we  have  V—  9  =  3  V^l  , 

and  V~=l=2V~l. 

Therefore,  V—  9X  V-4=SV  —  1X2V  —  I 


2.  Multiply  1+  VI  by  1-  V. 

Ans.  2. 

3.  Multiply  VIS  byj/—2. 

4.  Multiply  5+2  V^S  by  2-  V^3. 


SECTION  XII. 


EQUATIONS  OF  THE  SECOND  DEGREE. 

(178.)  According  to  Art.  96,  quadratic  equations,  or  equa- 
tions of  the  second  degree,  are  those  in  which  the  highest  power 
of  the  unknown  quantity  is  a  square. 

Quadratic  equations  are  divided  into  two  classes. 

I.  Equations  which  involve  only  the  square  of  the  unknown 
quantity  and  known  terms.     These  are  called  pure  quadratics. 
Of  this  description  are  the  equations 

ax*=b;  3x*+12=I50-x\  &c. 

They  are  sometimes  called  quadratic  equations  of  two  terms, 
because,  by  transposition  and  reduction,  they  can  always  be 
exhibited  under  the  general  form 

ax*=b. 

II.  Equations  which  involve  both  the  square  and  the  first 
power  of  the  unknown  quantity,  together  with  a  known  term. 
These  are  called  affected  or  complete  quadratics.     Of  this  de- 
scription are  the  equations 

ax*+bx=c;  x*-  10*=7  ;  !|L|+|=8. 

They  are  sometimes  called  quadratic  equations  of  three 
terms,  because,  by  transposition  and  reduction,  they  can  al- 
ways be  exhibited  under  the  general  form 


PURE  QUADRATIC  EQUATIONS. 

(179.)  The  equation 


138          EQUATIONS  OP  THE  SECOND  DEGREE. 

is  easily  solved.     Dividing  each  member  by  a,  it  becomes 

b 


Whence  x  =: 

If  -  be  a  particular  number,  either  integral  or  fractional,  we 
a 

can  extract  its  square  root  either  exactly  or  approximately  by 
the  rules  of  arithmetic. 

It  is  to  be  remarked,  that  since  the  square  both  of  +m  and 

— m  is  +7?ia,  so,  in  like  manner,  the  square  of  -f-  v/-  and  that 

of  —  y  -  are  both  +-.  Hence  the  above  equation  is  suscep- 
tible of  two  solutions,  or  has  two  roots ;  that  is,  there  are  two 
quantities  which,  when  substituted  for  x  in  the  original  equation, 
will  render  the  two  members  identical.  These  are 


For,  substituting  each  of  these  values  in  the  original  equation 
ax*=b,  it  becomes 

aX(+y  -)  =b,  or  aX-=b  ;  i.  e.,  b=b, 
and  «X(—  Y  -J  =b,  or  aX~=b  ;  i.  e.,  b=b. 


'"••,'.  H"  CO  1 
EXAMPLE   I. 

Find  the  values  of  x  which  satisfy  the  equation 

4^-7=3^+9. 

Transposing  terms,  4x2—  3#a=9+7. 
Reducing,  *a=16. 

Extracting  the  square  root, 


Hence  the  two  values  of  x  are  +4  and  —4,  and  they  may 
both  be  verified  by  substitution  in  the  original  equation. 
Thus,  taking  the  first  value,  we  have 


EQUATIONS  OF  THE  SECOND  DEGREE.  139 

4X(+4)a-7=3X(+4)a+9, 
or  4X16-7=3X16+9; 

that  is,  57=57. 

Taking  the  second  value  of  x,  we  have 

4X  (-4)a-7=3X  (-4)3+9, 

or  4X16-7=3X16+9,  as  before. 

From  the  preceding  examples  we  deduce  the  following 

RULE. 

Reduce  the  equation  to  the  form  axa=b ;  then  divide  by  the 
coefficient  of  xa,  and  extract  the  square  root  of  both  members  of 
the  equation. 

Ex.  2.  Given  za-17=130-2a;3,to  find  the  values  of  x. 

By  transposition,  3#3=147; 

therefore,  xa=49, 

and  x  =±7. 

Ex.  3.  Given  x*+ab=5x*9  to  find  the  values  ofx. 

By  transposition,  ab=4x* ; 

therefore,  ±  Vab=2x, 

and  x=- 


2a* 


Ex.  4.  Given  x+  -v/a3+xa=~7~7" :,  to  find  the  values  of  x. 

Va  +x 

Clearing  of  fractions,  we  obtain  a:Vaa+^a+aa+«a=2aa. 

By  transposition,       xVa^+ 

Squaring  both  sides,  aV+#4=a4— 
therefore,  3aV=a4, 

and  3a:a=aa ; 

whence  #a=-^; 

o 

therefore,  x = ± — -. 

v/3 

Ex.  5.  Given  — - — =45,  to  find  the  values  of  a:, 
y 

Ex.  6.  Given  a#2— 5c=faa— 3c+d,  to  find  the  values  of  X. 


140          EQUATIONS  OF  THE  SECOND  DEGREE. 

x*  5  7  299 

Ex.  7.  Given  -^—3+—x*=— -.—  xa+— — ,  to  find  the  values 
3  12        24  24 

of*. 

Ex.  8.  Given  —  :=:^  to  ^d  tne  values  of*. 

x 


Ans.  *=db3. 

Clearing  of  fractions  and  transposing,  we  find  in  each  mem- 
ber of  this  equation  a  binomial  factor,  which  being  canceled, 
the  equation  is  easily  solved. 

PROBLEMS. 

Prob.  1.  What  two  numbers  are  those  whose  sum  is  to  the 
greater  as  10  to  7 ;  and  whose  sum,  multiplied  by  the  less,  pro- 
duces 270  ? 

Let  10*=  their  sum. 

Then  7*=  the  greater  number, 

and  3*=  the  less. 

Whence  30*2=270, 

and  *2=9; 

therefore,  #=±3, 
and  the  numbers  are  ±21  and  ±9. 

Prob.  2.  What  two  numbers  are  those  whose  sum  is  to  the 
greater  as  m  to  n;  and  whose  sum,  multiplied  by  the  less,  is 
equal  to  a  ? 


,      an3  /a(m—n) 

Ans.  rb\/— 1  and  ±\/— -• 

m(m—n)  V        m 

Prob.  3.  What  number  is  that,  the  third  part  of  whose 
square  being  subtracted  from  20,  leaves  a  remainder  equal 
to  8? 

Prob.  4.  What  number  is  that,  the  mth  part  of  whose  square 
being  subtracted  from  a,  leaves  a  remainder  equal  to  b  ? 

Ans.  ±  Vm(a—b). 

12  3 

Prob.  5.  Find  three  numbers  in  the  ratio  of  -,  -,  and  -,  the 

sum  of  whose  squares  is  724. 


EQUATIONS    OF    THE    SECOND    DEGREE.  141 

Prob.  6.  Find  three  numbers  in  the  ratio  of  m,  n,  and  jo,  the 
sum  of  whose  squares  is  equal  to  a. 
Ans. 


an 

v  JSTH?+P;  and 


ap* 


Prob.  7.  Divide  the  number  49  into  two  such  parts,  that  the 
quotient  of  the  greater  divided  by  the  less,  may  be  to  the  quo- 
tient of  the  less  divided  by  the  greater,  as  £  to  f  . 

Ans.  21  and  28. 

Prob.  8.  Divide  the  number  a  into  two  such  parts,  that  the 
quotient  of  the  greater  divided  by  the  less,  may  be  to  the  quo- 
tient of  the  less  divided  by  the  greater,  as  m  to  n. 


Ans.  -  •  -  and 


Prob.  9.  There  are  two  square  grass  plats,  a  side  of  one  of 
which  is  10  yards  longer  than  a  side  of  the  other,  and  their 
areas  are  as  25  to  9.  What  are  the  lengths  of  the  sides  ? 

Prob.  10.  There  are  two  squares  whose  areas  are  as  m  to  n, 
and  a  side  of  one  exceeds  a  side  of  the  other  by  a.  What  are 
the  lengths  of  the  sides  ? 

a*/m 

Ans.  and 


Prob.  11.  Two  travelers,  A  and  B,  set  out  to  meet  each  other, 
A  leaving  Hartford  at  the  same  time  that  B  left  New  York. 
On  meeting,  it  appeared  that  A  had  traveled  18  miles  more 
than  B,  and  that  A  could  have  gone  B's  journey  in  15£  hours, 
but  B  would  have  been  28  hours  in  performing  A's  journey. 
What  was  the  distance  between,  Hartford  and  New  York  ? 

Ans.  126  miles. 

Prob.  12.  From  two  places  at  an  unknown  distance,  two 
bodies,  A  and  B,  move  toward  each  other,  A  going  a  miles  more 
than  B.  A  would  have  described  B's  distance  in  n  hours,  and 
B  would  have  described  A's  distance  in  m  hours.  What  was 
the  distance  of  the  two  places  from  each  other  ? 

Ans.  aX 
Prob.  13.  A  vintner  draws  a  certain  quantity  of  wine  out  of 


142  EQUATIONS  OF  THE  SECOND  DEGREE 

a  full  vessel  that  holds  256  gallons ;  and  then,  filling  the  vessel 
with  water,  draws  off  the  same  quantity  of  liquor  as  before, 
and  so  on  for  four  draughts,  when  there  were  only  81  gallons 
of  pure  wine  left.  How  much  wine  did  he  draw  each  time  ? 

Ans.  64,  48,  36,  and  27  gallons. 

Prob.  14.  A  number  a  is  diminished  by  the  nth  part  of  it- 
self, this  remainder  is  diminished  by  the  nth  part  of  itself,  and 
so  on  to  the  fourth  remainder,  which  is  equal  to  b.  Required 
the  value  of  n. 

Va 
Ans-   Va-VV 

Prob.  15.  Two  workmen,  A  and  B,  were  engaged  to  work 
for  a  certain  number  of  days  at  different  rates.  At  the  end  of 
the  time,  A,  who  had  played  4  of  those  days,  received  75  shil- 
lings, but  B,  who  had  played  7  of  those  days,  received  only 
48  shillings.  Now  had  B  only  played  4  days,  and  A  played  7 
days,  they  would  have  received  the  same  sum.  For  how 
many  days  were  they  engaged  ? 

Ans.  19  days. 

Prob.  16.  A  person  employed  two  laborers,  allowing  them 
different  wages.  At  the  end  of  a  certain  number  of  days,  the 
first,  who  had  played  a  days,  received  m  shillings,  and  the 
second,  who  had  played  b  days,  received  n  shillings.  Now  if 
the  second  had  played  a  days,  and  the  other  b  days,  they 
would  both  have  received  the  same  sum.  For  how  many  days 
were  they  engaged  ? 

b<Jm—a</n  . 
Ans.  days. 

—  J 


COMPLETE  QUADRATIC  EQUATIONS. 

(180.)  Suppose  we  have  the  equation 


in  -which  it  is  required  to  find  the  value  of  x. 

Since  each  member  of  the  equation  is  a  complete  square,  if 
we  extract  the  square  root,  we  shall  obtain  a  new  equation 
involving  only  the  first  power  of  x,  which  may  be  easily 
solved. 


EdUATIONS  OP  THE  SECOND  DEGREE.  143 

We  thus  have  x—  3=±1, 

and,  by  transposition, 

x=3±I=4,  or2. 

In  order  to  verify  these  values,  substitute  each  of  them  in 
place  of  x  in  the  given  equation.  Taking  the  first  value,  we 
shall  have 

4a-6x4+9=l; 
that  is,  16—24+9=1,  an  identical  equation. 

Taking  the  second  value  of  #,  we  obtain 

2'-6X2+9=l; 
that  is,  4—12+9=1,  an  identical  equation. 

Hence  we  see  that  a  complete  quadratic  equation  is  readily 
solved,  provided  each  member  of  the  equation  is  a  perfect 
square.  But  equations  seldom  occur  under  this  form.  Take, 
for  example, 

xa_6x=-8. 

The  preceding  method  seems  to  be  inapplicable,  because  the 
first  member  is  not  a  complete  square.  We  may,  however, 
render  it  a  complete  square  by  the  addition  of  9,  which  must 
also  be  added  to  the  second  member  to  preserve  the  equality. 
The  equation  thus  becomes 

x*—  Qx+9=9—  8=1, 
which  is  the  equation  first  proposed. 

The  peculiar  difficulty,  then,  in  resolving  complete  equa- 
tions of  the  second  degree,  consists  in  rendering  the  first  mem- 
ber an  exact  square. 

(181.)  In  order  to  discover  a  general  method  of  solution,  let 
us  take  the  equation 


which  is  the  general  form  of  equations  of  the  second  degree. 
We  begin  by  dividing  both  members  by  a,  the  coefficient  of  x*. 
The  equation  then  becomes 


For  convenience,  let  us  put  »=-  and  q=-;  we  shall  then 

a  a 

have 

x*+px=q. 

7* 


144  EQUATIONS    OF    THE    SECOND    DEGREE. 

We  have  see*n  that  if  we  can  by  any  transformation  render 
the  first  xmember  of  this  equation  the  perfect  square  of  a  bino- 
mial, we  can  reduce  the  equation  to  one  of  the  first  degree  by 
extracting  the  square  root. 

Now  we  know  that  the  square  of  a  binomial,  x+a,  or  a;2+ 
2ax+a*,  is  composed  of  the  square  of  the  first  term,  plus  twice 
the  product  of  the  first  term  by  the  second,  plus  the  square  of 

the  second  term. 

• 

Hence,  considering  x*+px  as  the  first  two  terms  of  the 
square  of  a  binomial,  and,  consequently,  px  as  being  twice  the 
product  of  the  first  term  of  the  binomial  by  the  second,  it  is 

evident  that  the  second  term  of  this  binomial  must  be    ,  for 


In  order,  therefore,  that  the  expression  x*  +px  may  be  ren- 
dered a  perfect  square,  we  must  add  to  it  the  square  of  this 

second  term  ^  ;  that  is,  the  square  of  half  the  coefficient  of  the 
first  power  of  x  ;  it  thus  becomes 


which  is  the  square  of  #+£•    But  since  we  have  added  ^-  to 

the  left-hand  member  of  the  equation,  in  order  that  the  equality 
may  not  be  destroyed,  we  must  add  the  same  quantity  to  the 
right-hand  member  also  ;  the  equation  thus  transformed  will 
become 


Extracting  its  square  root,  we  have 


Whence 

We  prefix  the  double  sign  ±,  because  the  square  both  of 


EaUATIONS    OF    THE    SECOND    DEGREE.  14" 


4   *s  +  (<7+-T')»  and  ever 


quadratic  equation  must  therefore  have  two  roots. 

(182.)  From  the  preceding  discussion  we  deduce  the  follow- 
ing general 

RULE  FOR  THE  SOLUTION  OF  A  COMPLETE  QUADRATIC  EQUATION 

1.  Transpose  all  the  known  quantities  to  one  side  of  the  equa- 
tion, and  all  the  terms  involving  the  unknown  quantity  to  th: 
other  side,  and  reduce  the  equation  to  the  form  axa+bx=c. 

2.  Divide  each  side  of  the  equation  by  the  coefficient  of  x2,  and 
add  to  each  member  the  square  of  half  the  coefficient  of  the  first 
power  of  x. 

3.  Extract  the  square  root  of  both  sides,  and  the  equation  will 
be  reduced  to  one  of  the  first  degree,  which  may  be  solved  in  the 
usual  manner. 

EXAMPLE    1. 

Solve  the  equation  x*—  I0x=  —  16. 

Completing  the  square  by  adding  to  each  member  the  square 
of  half  the  coefficient  of  the  second  term,  we  have 

x3-  10z+25=25-  16=9. 
Extracting  the  root,     x—  5=  ±3. 
Hence  #=5±3. 


An,.    \  *=*+»  -*• 
(  #=5—3=2. 


Thus  x  has  two  values,  either  8  or  2.     To  verify  them,  sub- 
stitute in  the  original  equation,  and  we  shall  have 

82-10X8=-16,  {.  e.,  64-80=-16  ; 
also,  23-10X2=-16,  z.  e.,    4-20=-16, 

both  of  which  are  identical  equations. 

EXAMPLE    2. 

Solve  the  equation         a:3 +62;=  —  8. 
Completing  the  square,  #3+6a:+9=9— 8=1. 
Extracting  the  root,       x  +3=±l. 
Hence  x  =— 3±1. 

Ans. 


146          EdUATIONS  OF  THE  SECOND  DEGREE. 

Proof.     (-2)3+6X-2=-8,i.  e.,    4-12=-8; 
also,  (_4)8+6X-4=-8,  f.  e.,  16-24= -8. 

Hence  x  has  two  values,  both  negative.  In  verifying  them, 
it  is  to  be  observed,  that  the  square  of  —2  is  +4,  and  —2  mul- 
tiplied by +6  gives  -12. 

EXAMPLE    3. 

Solve  the  equation          #a+6x=27. 
Completing  the  square,  #2+6z+9=27+9=36. 
Extracting  the  root,       x  +3= ±6. 
Hence  x  = — 3±6=  +3,  or  —9. 


EXAMPLE  4. 

Solve  the  equation  x9— 2x=24. 
Here  x  =  l±5=+6,  or  -4. 

EXAMPLE    5. 

Solve  the  equation        x3 — 8x  =  —  1 8. 
Completing  the  square,  or1 — 8#+16=16— 18=— 2. 

Hence  x  =  4±  V  —  2. 

Here  both  values  ofx  are  imaginary. 

EXAMPLE  6. 

¥. ret    :=<•£  —  -y      ,  N"'O'!  ^;'t. 

Solve  the  equation         #2— 6x=— 9. 

Completing  the  square,  #3— 6x4-9=9— 9=0. 

Extracting  the  root,       x  —  3=±0. 

Hence  x  =3±0. 

Here  the  two  values  of  x  are  equal  to  each  other. 

Ex.  7.  Given  2#a+ Sx— 20=70,  to  find  the  values  of  a:. 

Ans.  #=5,  or  —9. 

Ex.  8.  Given  3#q— 3#+6=5£,  to  find  the  values  of  x. 

Ans.  x=\9  or  i. 

(183.)  The  Rule  given  on  page  145  for  solving  a  complete 
quadratic  equation  is  applicable  to  all  cases ;  nevertheless,  a 
modification  of  this  method  is  sometimes  preferable. 

The  object  is  to  render  the  first  member  of  the  equation  a 
perfect  square.  After  the  equation  has  been  reduced  to  the 
form 


EdUATIONS    OP    THE    SECOND    DEGREE.  147 

the  square  may  be  completed  by  multiplying  the  equation  by 
four  times  the  coefficient  of  xa,  and  adding  to  both  sides  the 
square  of  the  coefficient  of  x. 
Thus  the  above  equation  multiplied  by  4#  becomes 

4#  V  +  4abx  =  4ac. 
Adding  6a  to  both  members,  we  have 

4a*x*+4abx+b*=4ac+b*. 
Extracting  the  square  root, 

2ax+b=±V4ac+b\ 
Transposing  b  and  dividing  by  2a, 


which  is  the  same  result  as  would  be  obtained  by  the  formei 
Rule  ;  but  by  this  new  method  we  have  avoided  the  introduc- 
tion of  fractions  in  completing  the  square. 

When  the  coefficient  of  x*  is  unity,  the  above  Rule  becomes, 
Multiply  the  equation  by  four,  and  add  to  each  member  the 
square  of  the  coefficient  of  x. 

Either  of  these  methods  of  completing  the  square  may  be 
practiced  at  pleasure  ;  but  the  first  method  is  to  be  preferred,  ex- 
cept when  its  application  would  involve  inconvenient  fractions. 

Ex.  9.  Given  i#a—  Jx+20J=42|,  to  find  the  values  of  x. 

Ans.  x—7,  or  —  6|. 

Ex.  10.  Given  #a-a;-40=170,  to  find  the  values  of  x. 

Ans.  #=15,  or  —14. 

Ex.  11.  Given  3za+2£-9=76,  to  find  the  values  of  x. 

Ex.  12.  Given  l#3-ix+7f  =8,  to  find  the  values  of  x. 


Ex.  13.  Given  3x  —  |—  -  —  *~~     =2,  to  find  the  values 
<&x  "~~  L       y  ~~  £x 

of*. 

We  must  first  clear  this  equation  of  fractions,  which  is  done 
by  multiplying  by  the  denominators  ;  we  thus  obtain 


8:ca--  18. 

Here  the  two  terms  containing  x3  balance  each  other,  and, 
uniting  similar  terms,  we  obtain 

8a;9-124z=-368. 


148  EatJATIONS    OF    THE   SECOND    DEGREE. 

Dividing  by  8,  we  have 
#a 
Completing  the  square, 


Extracting  the  root,  x  —  —  =±—  . 

31      15 

Hence  #=-7  -±—  -=114-,  or  4.  Ans. 

4      4 

90       90         27 

j&e.  14.  Given  ---  —  -  --  rr:=0,  to  find  the  values  of  a:. 
x     x+l    x+2 

(184.)  The  preceding  rules  will  enable  us  to  solve  not  only 
quadratic  equations,  but  all  equations  which  can  be  reduced  to 
the  form 

x*n+pxn=q; 

that  is,  all  equations  which  contain  only  two  powers  of  the  un 
known  quantity,  and  in  which  one  of  the  exponents  is  double  of 
the  other. 

For  if,  in  the  above  equation,  we  assume  y=xn,  then  y*=x*n, 
and  it  becomes 

y*+py=q. 

Solving  this  according  to  the  rule,  we  find 


2 

Extracting  the  nth  root  of  both  sides, 


*=v-f±vHr 


EXAMPLE    1. 

Given  x'—25x*=  — 144,  to  find  the  values  of  a?. 
Assuming  za=y,  the  above  becomes 

2/3-25y=-144. 

Whence  y=     16,  or  9. 

But,  since  x*=y,  x=±\/y. 

Therefore,  #=±  v/16,  or  ±  v/9. 

Thus  x  has  four  values,  viz.,  +4,  —4,  +3,  —3. 


EQUATIONS  OF  THE  SECOND  DEGREE.          140 

• 

To  verify  these  values : 

1st  value,  (+4)4-25X(+4)a=-144,  i.  e.,  256-400= -144. 
2d  value,  (-4)4-25X(-4)a=-144,  i.  e.,  256-400^-144. 
3d  value,  (+3)4-25X(+3)a=-144,  i.  e.,  81 -225= -144. 
4th  value,  (-3)4-25X(-3)8=-144,  i.  e.,  81 -225= -144. 

EXAMPLE    2. 

Given  x*— 7#a=8,  to  find  the  values  of  x. 

Assuming  x*=y,  we  have 

y'-7y=8. 
Whence  y  =8,  or  —  1. 

Therefore,  #=±  -y/8,  or  ±  V  —  1,  the  two  last  of  which  roots 
are  imaginary. 

EXAMPLE  3. 

Given  x6— 2#8=48,  to  find  the  values  of  x. 

Assuming         z*—y,  the  above  becomes 

y— 2y=4S. 
Whence  y  =8,  or  —  6. 

And  since  x3=y,  therefore  x—  tyy. 

Hence  two  of  the  roots  of  the  above  equation  are  2  and—  V6. 

This  equation  has  four  other  roots,  which  can  not  be  de- 
termined by  this  process. 

EXAMPLE  4. 

Given  2x— <7^/x=99t  to  find  the  values  of  #. 
Assuming  ^/x=y9  this  equation  becomes 
2ys-7y=99. 

Whence  y=9,  or  — —. 

And  since  </x=y,  therefore  x—y^. 

„„  121 

Whence  x =81,  or  — j-% 

Although  the  square  root  of  81  is  generally  ambiguous,  and 
may  be  either  +9  or  —9,  still,  in  verifying  the  preceding 
values,  v/x  can  not  be  taken  equal  to  —9,  because  81  was  ob- 
tained by  multiplying  +9  by  itself.  For  a  like  reason,  ^/x 

can  not  be  taken  equal  to  +•«••     A  similar  remark  is  applica- 


150          EaUATIONS  OF  THE  SECOND  DEGREE. 

• 

ble  to  Exs.  13  and  14  on  the  next  page,  and  also  to  Ex.  7, 
page  186. 

.£  | 

EXAMPLE    5. 


Given  Vx+12+Vx+12=6,  to  find  the  values  of  x. 
Assuming  x+12=y,  this  equation  becomes 

which  evidently  belongs  to  the  same  class  as  the  previous  ex- 
amples.    Completing  the  square,  we  shall  have 

y*=2,  or  —3. 
Raising  both  sides  of  the  equation  to  the  fourth  power, 

y=16,  or  81. 
Therefore,  x  or  (y— 12) =4,  or  69. 

EXAMPLE    6. 

Given  2#a-fV2£a+l  =  ll,  to  find  the  values  ofx. 
Adding  1  to  each  member  of  the  equation,  it  becomes 


Assuming  2#a + 1  =y,  then 

!_ 
y-ry  — i*. 

Completing  the  square,  we  find 

y*=3,or-4; 
that  is,  V2za+l=3,  or  -4. 

Therefore,  2za+l=9,  or  16, 

and  a;a=4,  or  — . 

Hence  x=+2,  -2,  +\/y»  ""Va" 

It  may  be  remarked,  that  in  equations  of  this  kind  it  is  gen- 
erally unnecessary  to  substitute  a  new  letter,  y,  which  has  been 
done  in  the  preceding  examples  simply  for  the  sake  of  per- 
spicuity. 

Ex.  7.  Given  #4+4za=12,  to  find  the  values  of  #. 

Ans.  x=±V2,or±V^6 


EQUATIONS  OF  THE  SECOND  DEGREE.          151 

Ex.  8.  Given  #6—  8#3=513,  to  find  the  values  of  #. 

Ans.  #=3,  or  —    ~ 

6  3 

Ex.  9.  Given  xs-\-xs=75Q1  to  find  the  values  of  x. 

Ans.  #=243,  or  — 

Ex.  10.  Given  |#6—  i#3=—  ?V»  to  &*&  one  value  of  #. 

Ans.  #= 

Ex.  11.  Given  &c*+3aA=2,  to  find  the  values  of  #. 

.Aws.  #=|,  or  —8. 
Ex.  12.  Given  a4—  12#3+44#8—  48#=9009,  to  find  the  values 


This  equation  may  be  expressed  as  follows  : 
(#a-6#)a+8(#3-6#)=9009. 


Ans.  #=13,  or  —7,  or  3±  •/  —  90. 

.  13.  Given  \x—  £x/#=22£,  to  find  the  values  of  x. 

361 

Ans.  #=49,  or  —  —  . 
y 


Ex.  14.  Given  VlO+#-  VlO+#=2,  to  find  the  values  of  x. 

Ans.  #=6,  or  —9. 
Ex.  15.  Given  #8+20#8— 10=59  to  find  the  values  of  x. 

Ans.  #=V3,  or  V— 23. 
Ex.  16.  Given  3#an— 2#n+3=ll,  to  find  the  values  of  x. 

Ans.  x=  V2,  or  V^. 
Ex.  17.  Given  #a— aV3=#— 1-/3,  to  find  the  values  of  x. 

/3-1 

,  or 


.  18.  Given  V  1  +#-#'-  2  (l+#-#2)=i,  to  find  the  values 


Ans.  i±a  V4,  or 

(185.)  We  have  seen  that  every  equation  of  the  second  de- 
gree has  two  roots,  or  that  there  are  two  quantities  which,  when 
ubstituted  for  x  in  the  original  equation,  will  render  the  two 
nembers  identical.  In  like  manner,  we  shall  find  that  every 
Aquation  of  the  third  degree  has  three  roots  ;  an  equation  of  the 
burth  degree  has/owr  roots  ;  and,  in  general,  an  equation  has 
is  many  roots  as  it  has  dimensions. 


152  PROBLEMS    PRODUCING    ClUADRATIC    EdUATIONS. 

Before  determining  the  degree  of  an  equation,  it  should  be 
freed  from  fractions,  from  negative  exponents,  and  from  the 
radical  signs  which  affect  its  unknown  quantities.  Examples 
4,  5,  13,  and  14  are  thus  found  to  furnish  equations  of  the  sec- 
ond degree,  while  examples  6  and  18  furnish  equations  of  the 
fourth  degree. 

The  above  method  of  solving  the  equation  z*n+pxn=q  will 
not  always  give  us  all  of  the  roots,  and  we  must  have  recourse 
to  different  processes  to  discover  the  remaining  roots.  The 
subject  will  be  resumed  in  Section  XX. 


PROBLEMS  PRODUCING  QUADRATIC  EQUATIONS. 

Prob.  1.  It  is  required  to  find  two  numbers,  such  that  their 
difference  shall  be  8,  and  their  product  240. 

Let  x  =  the  least  number. 

Then  will  x+S  =  the  greater. 

And  by  the  question  x(x+8)=x*+8x=240. 

Therefore,  x=12,  the  less  number, 

#+8=20,  the  greater. 

Proof.  20-12=8,  the  first  condition. 

20X12=240,  the  second  condition. 

Prob.  2.  The  Receiving  Reservoir  at  Yorkville  is  a  rectan- 
gle, 60  rods  longer  than  it  is  broad,  and  its  area  is  5500  square 
rods.  Required  its  length  and  breadth  ? 

Prob.  3.  What  two  numbers  are  those  whose  difference  is 
2<z,  and  product  b  1 

Ans.  «=fc\/<za+&,  and  —  a±Va*+b. 

Prob.  4.  It  is  required  to  divide  the  number  60  into  two  such 
parts  that  their  product  shall  be  864. 

Let  x  =  one  of  the  parts. 

Then  will  60 -a:  =  the  other  part. 

And  by  the  question,  #(60— x)=60x— a8 =864. 

The  parts  are  36  and  24.  Ans. 

Prob.  5.  In  a  parcel  which  contains  52  coins  of  silver  and 
copper,  each  silver  coin  is  worth  as  many  cents  as  there  are 
copper  coins,  and  each  copper  coin  is  worth  as  many  cents  as 
there  are  silver  coins,  and  the  whole  are  worth  two  dollars 
How  many  are  there  of  each  ? 


PROBLEMS    PRODUCING    QUADRATIC    EQUATIONS.  153 

Prob.  6.  What  two  numbers  are  those  whose  sum  is  2a,  and 
product  b? 

Ans.  a+  Vcf—b,  and  a—  Va*—b. 

Prob.  7.  There  is  a  number  consisting  of  two  digits  whose 
sum  is  10,  and  the  sum  of  their  squares  is  58.  Required  the 
number. 

Let  x  =  the  first  digit. 

Then  will  10—  x  =  the  second  digit. 

And  x*+(W—x)*=2x*—  20^+100=58. 

That  is,          x*-Wx=-21, 


x=5±2=7,  or  3. 
Hence  the  number  is  73,  or  37. 

The  two  values  of  x  are  the  required  digits  whose  sum  is 
10.  It  will  be  observed  that  we  put  x  to  represent  the  first 
digit,  whereas  we  find  it  may  equal  the  second  as  well  as  the 
first.  The  reason  is,  that  we  have  here  imposed  a  condition 
which  does  not  enter  into  the  equation.  If  x  represent  either 
of  the  required  digits,  then  10—  x  will  represent  the  other,  and 
hence  the  values  of  x  found  by  solving  the  equation  should 
give  both  digits.  Beginners  are  very  apt  thus,  in  the  state- 
ment of  a  problem,  to  impose  conditions  which  do  not  appear 
in  the  equation. 

The  preceding  example,  and  all  others  of  the  same  class, 
may  be  solved  without  completing  the  square.  Thus, 

Let  x  represent  the  half  difference  of  the  two  digits. 

Then,  according  to  the  principle  on  page  67, 
5+x  will  represent  the  greater  of  the  two  digits, 
5-x  "  the  less  " 

The  square  of  5+x  is  25+10#+  x3, 
5-x      25-10:r+  x\ 

The  sum  is  50          +2#a,  which,  according  to  the 

problem,  =58. 

Hence  2x*=  8, 

or  x't=  4, 

and  x  =  ±2. 

Therefore,  5+x  =7,  the  greater  digit, 

5  —  x  =3,  the  less  digit. 


154  PROBLEMS  .PRODUCING    dUADRATIC    ECIUATIONS. 

Prob.  8.  Find  two  numbers  such  that  the  product  of  their 
sum  and  difference  may  be  5,  and  the  product  of  the  sum  of 
their  squares  and  the  difference  of  their  squares  may  be  65. 

Prob.  9.  Find  two  numbers  such  that  the  product  of  their 
sum  and  difference  may  be  «,  and  the  product  of  the  sum  of 
their  squares  and  the  difference  of  their  squares  may  be  ma. 

Ans. 

Prob.  10.  A  laborer  dug  two  trenches,  whose  united  length 
was  26  yards,  for  356  shillings,  and  the  digging  of  each  of 
them  cost  as  many  shillings  per  yard  as  there  were  yards  in 
its  length.  What  was  the  length  of  each  ? 

Ans.  10,  or  16  yards. 

Prob.  11.  What  two  numbers  are  those  whose  sum  is  2a, 
and  the  sum  of  their  squares  is  2b  ? 

Ans.  a-\-  Vb—cf,  and  a—  Vb—a*. 

Prob.  12.  A  farmer  bought  a  number  of  sheep  for  80  dollars, 
and  if  he  had  bought  four  more  for  the  same  money,  he  would 
have  paid  one  dollar  less  for  each.  How  many  did  he  buy  ? 

Let  x  represent  the  number  of  sheep. 

80 
Then  will  --  be  the  price  of  each. 

80 
And  --T—  would  be  the  price  of  each,  if  he  had  bought  four 


more  for  the  same  money. 
But  by  the  question  we  have 
80       80 


Solving  this  equation,  we  obtain 

x  =16.  Ans. 

Prob.  13.  A  person  bought  a  number  of  articles  for  a  dol- 
lars. If  he  had  bought  2b  more  for  the  same  money,  he  would 
have  paid  c  dollars  less  for  each.  How  many  did  he  buy  ? 

Prob.  14.  It  is  required  to  find  three  numbers  such  that  the 


PROBLEMS    PRODUCING    aUADRATIC    EQUATIONS.  155 

product  of  the  first  and  second  may  be  15,  the  product  of  the 
first  and  third  21,  and  the  sum  of  the  squares  of  the  second  and 
third  74. 

Ans.  3,  5,  and  7. 

Prob.  15.  It  is  required  to  find  three  numbers  such  that  the 
product  of  the  first  and  second  may  be  a,  the  product  of  the 
first  and  third  b,  and  the  sum  of  the  squares  of  the  second  and 
third  c. 


Prob.  16.  The  sum  of  two  numbers  is  16,  and  the  sum  of 
their  cubes  1072.     What  are  those  numbers  ? 

Ans.  7  and  9. 

Prob.  17.  The  sum  of  two  numbers  is  2a,  and  the  sum  of 
their  cubes  is  2b.     What  are  the  numbers  ? 


/b-a*  / 

Ans.  a+y  — — and  «—  w- 


b— a* 


Sa  V     Sa 

Prob.  18.  Two  magnets,  whose  powers  of  attraction  are  as 
4  to  9,  are  placed  at  a  distance  of  20  inches  from  each  other. 
It  is  required  to  find,  on  the  line  which  joins  their  centers,  the 
point  where  a  needle  would  be  equally  attracted  by  both,  ad- 
mitting that  the  intensity  of  magnetic  attraction  varies  inverse- 
ly as  the  square  of  the  distance. 

,         (  8  inches  from  the  weakest  magnet, 

(  or  —40  inches  from  the  weakest  magnet. 

Prob.  19.  Two  magnets,  whose  powers  are  as  m  to  n,  are 
placed  at  a  distance  of  a  feet  from  each  other.  It  is  required 
to  find,  on  the  line  which  joins  their  centers,  the  point  which  is 
equally  attracted  by  both. 

The  distance  from  the  magnet  m  is 
Ans. 

The  distance  from  the  magnet  n  is 


Prob.  20.  A  set  out  from  C  toward  D,  and  traveled  6  miles 
an  hour  After  he  had  gone  45  miles,  B  set  out  from  D  to- 
ward C,  and  went  every  hour  ^V  of  the  entire  distance ;  and 
after  he  had  traveled  as  many  hours  as  he  went  miles  in  one 


156  QUADRATIC  EQUATIONS 


. 


hour,  he  met  A.     Required  the  distance  between  the  places  C 
and  D. 

Ans.  Either  100  miles,  or  180  miles. 

Prob.  21.  A  set  out  from  C  toward  D,  and  traveled  a  miles 
per  hour.  After  he  had  gone  b  miles,  B  set  out  from  D  toward 
C,  and  went  every  hour  ^th  of  the  entire  distance ;  and  after 
he  had  traveled  as  many  hours  as  he  went  miles  in  one  hour, 
he  met  A.  Required  the  distance  between  the  places  C  and  D. 


Prob.  22.  By  selling  my  horse  for  24  dollars,  I  lose  as  much 
per  cent,  as  the  horse  cost  me.  What  was  the  first  cost  of 
the  horse  ? 

Ans.  40,  or  60  dollars. 


aUADEATIC  EaUATIONS  CONTAINING  TWO  UNKNOWN  QUAN- 
TITIES. 

(186.)  An  equation  containing  two  unknown  quantities  is 
said  to  be  of  the  second  degree  when  it  involves  terms  in  which 
the  sum  of  the  exponents  of  the  unknown  quantities  is  equal  to  2, 
but  never  exceeds  2.  Thus, 


and  7xy—  4x+y  =40, 

are  equations  of  the  second  degree. 

The  solution  of  two  equations  of  the  second  degree  contain- 
ing two  unknown  quantities,  generally  involves  the  solution  of 
an  equation  of  the  fourth  degree  containing  one  unknown  quan- 
tity. Hence  the  principles  hitherto  established  are  not  suffi- 
cient to  enable  us  to  solve  all  equations  of  this  description. 
Yet  there  are  particular  cases  in  which  they  may  be  reduced 
either  to  pure  or  affected  quadratics,  and  the  roots  determined 
in  the  ordinary  manner. 

(187.)  When  one  of  the  equations  is  a  simple  equation,  it  is 
generally  best  to  find  an  expression  for  the  value  of  one  of  the 
unknown  quantities  from  the  simple  equation,  and  substitute 
this  value  in  the  place  of  its  equal  in  the  other  equation.  The 
resulting  equation  will  be  of  the  second  degree,  and  may  be 
solved  by  the  ordinary  rules. 


CONTAINING    TWO    UNKNOWN    QUANTITIES.  157 

Ex.  1.  Given  x*+3xy—  y=23  )  to   find   the   values  of  x 

x  +2y  =  7  j      and  y. 
From  the  second  equation,  we  find 

x  =7—  2y. 

Whence  x*=49-28y+4y*. 

And,  substituting  this  value  in  the  first  equation,  we  have 


a  common  quadratic  equation,  which  may  be  solved  in  the 
usual  manner. 

Ans.  x=3,  and  y=2. 

Ex.  2.  Given  2x*+  xy—5y*=2Q  )  to  find  the  values  of  x 
2x  —  3y  =  I  )      and  y. 

Ans.  x=5,  y=3. 

Wx+y 
Ex.  3.  Given  —  —  £=a;       A    n    ,    ,         ,         c         , 

3  y  >  to  find  the  values  of  x  and  y. 


9y-9x=l8 

Ans.  x=2,  y=4. 

(188.)  When  the  same  algebraic  expression  is  involved  to 
different  powers,  it  is  sometimes  best  to  regard  this  expression 
as  the  unknown  quantity. 

Ex.  4.  Given  x*+2xy+y*  +2x=I20—2y  )  to  find  the  val- 
xy—  y*  =8  S      uesofa;andy. 

Here  the  first  equation  may  be  put  under  the  form 


where  x+y  may  be  regarded  as  a  single  quantity,  and,  by 
completing  the  square,  we  shall  find  its  value  to  be 
either  10,  or  —12. 

Proceeding  now  as  in  the  last  Article,  we  shall  find 

x=6,  or  9,  or  —  9qpv/5, 
y=4,  or  1,  or  —  3±v/5. 

Ex.  5.  Given  4xy=96—x*y*  ) 

y    I  to  find  the  values  of  x  and  y. 
x=Q 


Here  we  may  regard  xy  as  the  unknown  quantity,  and  we 
shall  find  its  value  from  the  first  equation  to  be 

either  8,  or  —12. 

Proceeding  again  as  in  the  former  Article,  we  shall  find 


158  QUADRATIC  EQUATIONS 

z=2,  or  4,  or  3±  </21, 
y=4,  or  2,  or 


#a    4x    85  \ 
.  6.  Given  -H  —  =-7-  f  x    ,,    ,  A,         ,         r         , 

y      y      9  >  to  find  the  values  of  #  and  y. 


x 
Here  -  may  be  treated  as  the  unknown  quantity,  and  we 

shall  find  its  value  to  be 

5  17 

either  g,  or  —  —  . 

From  which  we  find 


,_..    ,^^. 

(189.)  When  the  sum  of  the  dimensions  of  the  unknown 
quantities  is  the  same  in  every  term  of  the  two  equations,  it  is 
sometimes  best  to  substitute  for  one  of  the  unknown  quantities 
the  product  of  the  other  by  a  third  unknown  quantity. 

Ex.  7.  Given  x*+xy  =56 


xy  t0  find  the  ValueS  °f  x  and 

Here,  if  we  assume  x=vy,  we  shall  have 


From  the  first  of  these  equations, 

£Jfc 

1)  -f-V 

60 

and  from  the  second,          y8=~~T7;  •» 

v  "r~2 

.  60         56 

therefore,  — — = -r — . 

v+2    ir+t> 

From  which,  after  completing  the  square,  we  obtain 

4  7 

v=-,  or  --. 

o  O        ;  -at  ,}- 

Substituting  either  of  these  values  in  one  of  the  preceding 
expressions  for  ya,  we  shall  obtain  the  values  of  y ;  and  since 
x=vy,  we  may  easily  obtain  the  values  of  #. 

.^•=±14,  or 
An*.  \  -,        10>  or 


CONTAINING    TWO    UNKNOWN    QUANTITIES.  159 

Ex.  8.  Given  2x*+3xy+  y*=20  )  to  find  the  values  of  x 

5x*  +4y2=41  )      and  y. 
If  we  assume  x=vy,  we  shall  find 
_1        13 

whence,  as  before,  we  shall  obtain 

#=±1,  or     > 

±2 

'=±3,  or 
V. 

Ex.  9.  Given 


_  i  to  find  the  values  of  x  and  v. 
xy—y1  =12 

If  we  assume  x—vy9  we  shall  find 
_7        U 

whence,  as  before, 

±11 
:      fc7,  or  —^^ 

Ans.    .  ±g 

^±4'°r72 

(190.)  When  the  unknown  quantities  in  each  equation  are 
similarly  involved,  it  is  sometimes  best  to  substitute  for  the 
unknown  quantities  the  sum  and  difference  of  two  other  quan- 
tities, or  the  sum  and  product  of  two  other  quantities. 

Ex.  10.  Given  —+^-=18  ) 

y     x          >  to  find  the  values  of  x  and  y. 

x+y=12  ) 
Here  let  us  assume 

x=z+v, 
y=z—v. 

Then,  by  adding  these  two  equations  together,  we  shall  have 

x+y=2z=I2,  or  %=6; 
that  is,  x=6+v,  and  y=Q— v. 

But,  from  the  first  equation,  we  find 


8 


160  UUADRATIC    EQUATIONS 

Substituting  the  preceding  values  of  x  and  y  in  this  equa- 
tion, and  reducing,  we  obtain 

432+36u2=648-18va. 

Whence  t>=±2. 

Therefore,  #=4,  or  8, 

and  y=S,  or  4. 

Ex.  11.  Given  #6+V6=3368 
y 


to  find  the  values  of  x  and  y. 
x  +y=       8  ) 

(  x=3,  or  5, 

Ans.  ] 

(  y—  5,  or  3. 


-         .  • 

.r=5,  or  0, 


Ex.  12.  Given  x3  +y3  =341  ),<.,,.         ,          -          . 
a  a_         |  to  find  the  values  of  x  and 

x=5,  or  0 
y=6,  or  5. 


PROBLEMS. 

1.  Divide  the  number  100  into  two  such  parts,  that  the  sum 
of  their  square  roots  may  be  14. 

Ans.  64  and  36. 

2.  Divide  the  number  a  into  two  such  parts,  that  the  sum 
of  their  square  roots  may  be  b. 


3.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their  fourth 
powers  is  706.     What  are  the  numbers  ? 

Ans.  3  and  5. 

4.  The  sum  of  two  numbers  is  2#,  and  the  sum  of  their 
fourth  powers  is  2b.     What  are  the  numbers  ? 

Ans. 


5.  The  sum  of  two  numbers  is  6,  and  the  sum  of  their  fifth 
powers  is  1056.     What  are  the  numbers  ? 

Ans.  2  and  4. 

6.  The  sum  of  two  numbers  is  2a,  and  the  sum  of  their  fifth 
powers  is  b.     What  are  the  numbers  ? 


Ans.  a±yV—-+-^~a'. 


CONTAINING    TWO    UNKNOWN    QUANTITIES.  16f 

7.  What  two  numbers  are  those  whose  product  is  120  ;  and 
if  the  greater  be  increased  by  8  and  the  less  by  5,  the  product 
of  the  two  numbers  thus  obtained  shall  be  300  ? 

Ans.  12  and  10,  or  16  and  7.5. 

8.  What  two  numbers  are  those  whose  product  is  a ;  and 
if  the  greater  be  increased  by  b  and  the  less  by  c,  the  product 
of  the  two  numbers  thus  obtained  shall  be  d? 


m         /m*    ab          m         /m?    ab 

d—a—bc 

where  m=. . 

c 

9.  Find  two  numbers  such  that  their  sum,  their   product, 
and  the  difference  of  their  squares  may  be  all  equal  to  one  an- 
other. 

3         5,1         5 

Ans.  2+\/4»  and  2+v/4' 

that  is,  2.618,  and  1.618,  nearly. 

10.  Divide  the  number  100  into  two  such  parts,  that  their 
product  may  be  equal  to  the  difference  of  their  squares. 

Ans.  38.197,  and  61.803. 

11.  Divide  the  number  a  into  two  such  parts,  that  their  prod- 
uct may  ftfe  equal  to  the  difference  of  their  squares. 

Ans. and  — 


DISCUSSION   OP   THE    GENERAL   EQUATION  OP   THE   SECOND 
DEGREE. 

(191.)  We  have  seen,  Art.  181,  that  every  equation  of  the 
second  degree  may  be  reduced  to  the  form 

a;3  +px=q, 

where  p  and  q  represent  known  quantities,  either  positive  OP 
negative,  integral  or  fractional. 

The  value  of  x  in  this  equation  is 

either  x=-l 


or 


162  DISCUSSION    OF    THE 

And,  since  these  values  necessarily  result  from  the  general 
equation,  we  infer, 

astfi)  ^nsJiiturj  ow)  srh  V> 


PROPERTY  I. 

Every  equation  of  the  second  degree  has  two  roots,  and  only 
two. 

A  root  of  an  equation  is  such  a  number  as,  being  substituted 
for  the  unknown  quantity,  will  satisfy  the  equation. 

This  principle  has  been  often  exemplified  in  the  preceding 
pages.  Two  values  have  uniformly  been  found  for  x,  although 
both  values  may  not  be  applicable  to  the  problem  which  fur- 
nishes the  equation.  This  property  will  be  found  demon- 
strated in  a  general  manner  in  Art.  294. 

(192.)  If  we  multiply 


Id  ,OE 


we  shall  obtain  x*+px—q=Q, 

which  was  the  equation  originally  proposed. 

Tj 

' 

. 

PROPERTY  II. 

Every  equation  of  the  second  degree,  whose  roots  are  a  and  b, 
may  be  resolved  into  the  two  factors  x— a  and  x— b. 

Ex.  1.  Thus  the  equation 


may  be  resolved  into  the  factors   j  XJ^ 

where  8  and  2  are  the  roots  of  the  given  equation. 

It  is  also  obvious  that  if  a  is  a  root  of  an  equation  of  the  sec- 
ond degree,  this  equation  must  be  divisible  by  x—  a.  Thus 
the  preceding  equation  is  divisible  by  x—8,  giving  the  quotient 


Ex.  2.  The  roots  of  the  equation 

• 


GENERAL  EdUATION  OF  THE  SECOND  DEGEEE.      163 


are  —2  and  —4.  Resolve  it  into  its  factors. 
Ex.  3.  The  roots  of  the  equation 


are  +3  and  —9.     Resolve  it  into  its  factors. 

Ex.  4.  The  roots  of  the  equation 

x*— 2x— 24=0, 
are  +6  and  —4.     Resolve  it  into  its  factors. 

(193.)  If  we  add  together  the  two  values  of  x  in  the  gen- 
eral equation  of  the  second  degree,  the  radical  parts  having 
opposite  signs  disappear,  and  we  obtain 

"~2~2=~^' 
Hence, 

PROPERTY  III. 

The  algebraic  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  the  second  term  of  the  equation,  taken  with  a  contrary  sign. 

Thus,  in  Ex.  1,  page  145, 


the  two  roots  are  8  and  2,  whose  sum  is  +10,  the  coefficient 
of  x  taken  with  a  contrary  sign. 
In  the  equation 


the  two  roots  are  —2  and  —4. 

In  the  equation 


the  two  roots  are  —6  and  —10. 

If  the  two  roots  are  equal  numerically,  but  have  opposite 
signs,  their  sum  is  zero,  and  the  second  term  of  the  equation 
vanishes.  Thus  the  two  roots  of  the  equation  x2=16,  are  +4 
and  —4,  whose  sum  is  zero.  This  equation  may  be  written 


(194.)  If  we  multiply  together  the  two  values  of  #  (observ- 
ing that  the  product  of  the  sum  and  difference  of  two  quan- 
tities is  equal  to  the  difference  of  their  squares),  we  obtain 


164  DISCUSSION    OF    THE 


Hence, 

PROPERTY  IV. 

The  product  of  the  two  roots  is  equal  to  the  second  member  of 
the  equation,  taken  with  a  contrary  sign. 
Thus,  in  the  equation 


the  product  of  the  two  roots  8  and  2  is  +16,  which  is  equal  to 
the  second  member  of  the  equation  taken  with  a  contrary  sign. 
So,  also,  in  the  equation 


whose  two  roots  are  +3  and  —9,  their  product  is  —27. 

The  two  last  properties  enable  us  readily  to  form  an  equa- 
tion when  its  roots  are  known. 

Ex.  1.  Let  it  be  required  to  form  the  equation  whose  roots 
are  2  and  8. 

According  to  Property  III.,  the  coefficient  of  the  second 
term  of  the  equation  must  be  —10;  and,  from  Property  IV., 
the  second  member  of  the  equation  must  be  —16.  Hence  the 
equation  is 


Ex.  2.  Form  the  equation  whose  roots  are  3  and  5. 
Ex.  3.  Form  the  equation  whose  roots  are  —4  and  —7. 
Ex.  4.  Form  the  equation  whose  roots  are  5  and  —9. 
Ex.  5.  Form  the  equation  whose  roots  are  —6  and  +11. 

REAL  AND  IMAGINARY  VALUES  OF  THE  UNKNOWN  QUANTITY. 

(195.)  The  values  of  a:  in  the  general  equation  of  the  second 
degree  are 


Values  of  the  unknown  quantity  which  are  not  imaginary 
are,  for  the  sake  of  distinction,  called  real. 


GENERAL  EUUATION  OP  THE  SECOND  DEGREE.      165 

Since  ^-,  being  a  square,  is  positive  for  all  real  values  of  p, 

P* 

it  follows  that  the  expression  q+^  can  only  be  rendered  neg- 
ative by  the  sign  of  q. 

When  q  is  positive,  or  when  q  is  negative  and  numerically 

2?  59 

less  than  — ,  then  will  <7+^r  be  positive,  and,  consequently, 


\/q+  ~r  wnll  be  real.     This  happens  in  nearly  all  the  preced- 
ing examples. 

When  q  is  negative,  and  numerically  greater  than     -,  then 


q+*-r  will  be  negative,  and,  consequently,  y<7+^r  will  be  im- 
aginary.    This  happens  in  Ex.  5,  page  146. 


CASE   I. 


When         +  ~r  i 


1.  When,  in  the  equation  x*+px=q,  p  is  negative,  and  — 


is  numerically  greater  than  y^+x*  both  values  of  x  w;i7/  &e 

raz/  and  positive. 

This  happens  in  the  equation 

x*—  6x=-8, 
whose  two  roots  are  4  and  2. 

Also  in  the  equation 


whose  two  roots  are  8  and  2. 

f) 
2.  When  p  is  positive,  and       is  numerically  greater  than 


/       P" 
Y/  ?+ 4  >  ^;0^  values  of  x  wu7/  6e  real  and  negative. 


4 

This  happens  in  the  equation 


166  DISCUSSION    OF    THE 


whose  two  roots  are  —2  and  —4. 
Also  in  the  equation 


whose  two  roots  are  —6  and  —10. 

3.  When  |-  is  numerically  less  than  y  <7+  ~T»  both  values  of 
x  will  be  real,  the  one  positive  and  the  other  negative. 
This  happens  in  the  equation 


whose  roots  are  +3  and  —9. 
Also  in  the  equation 

' 
whose  roots  are  +6  and  —4. 


CASE   II. 

(196.)   When  y  q+^r  is  imaginary. 

In  this  case,  both  values  of  x  are  imaginary. 
This  happens  in  the  equation 

x*-8x=-18, 

whose  roots  are  4±V  — 2. 

We  will  now  prove  that  in  this  case  the  conditions  of  the 
question  are  incompatible  with  each  other,  and  therefore  the 
values  of  x  ought  to  be  imaginary.  The  demonstration  de- 
pends upon  the  following  principle : 

The  greatest  product  which  can  be  obtained  by  dividing  a 
number  into  two  parts  and  multiplying  them  together,  is  the 
square  of  half  that  number. 

Let  p  =  the  given  number, 
and      d  =  the  difference  of  the  parts. 

Then,  from  page  67,  ^-f-g  =  the  greater  part, 

p    d 

~~-  -  —  the  less  part, 


GENERAL  EQUATION  OF  THE  SECOND  DEGREE.      167 
»8   d3 

and  ^ — —  =  the  product  of  the  parts. 

Now,  since  p  is  a  given  quantity,  it  is  plain  that  this  expres- 

2 

sion  will  be  the  greatest  possible  when  d=0 ;  that  is,  ^-  is  the 

greatest  product,  which  is  the  square  of  ^,  half  the  given 

number. 

For  example,  let  12  be  the  number  to  be  divided. 
We  have  12=1  +  11;  and  11X1  =  11. 
12=2  +  10;  and  10X2=20. 
12=3+  9;  and  9X3=27. 
12=4+  8;  and  8X4=32. 
12=5+  7;  and  7X5=35. 
12=6+  6;  and  6x6=36. 

We  here  see  that  the  smaller  the  difference  of  the  two  parts, 
the  greater  is  their  product ;  and  this  product  is  greatest  when 
the  two  parts  are  equal. 
Now,  in  the  equation 

x*—px=-q, 
p  is  the  sum  of  the  two  roots,  and  q  is  their  product.     There- 

P* 
fore  q  can  never  be  greater  than  ~. 

If,  then,  any  problem  furnishes  an  equation  in  which  q  is 
negative,  and  greater  than  ^-,  we  infer  that  the  conditions  of 

the  question  are  incompatible  with  each  other.  • 
Thus,  in  the  example 

=^-=9,  which  is  numerically  less  than  q.  The  equation  re- 
quires us  to  divide  the  number  6  into  two  parts  whose  product 
shall  be  10,  which  is  an  impossibility;  and,  accordingly,  in 
solving  the  equation,  we  obtain  imaginary  values  for  x. 

Hence  an  imaginary  root  indicates  an  absurdity  in  the  pro- 
posed question  which  furnished  the  equation. 

Suppose  it  is  required  to  divide  8  into  two  such  parts  that 
their  product  shall  be  18. 


168  DISCUSSION    OF   PARTICULAR    PROBLEMS. 

Let  x  =  one  of "the  parts, 

and  S-x  =  the  other. 

Then,  by  the  conditions, 

^v  •~#)~- 

Whence  x*-8x=-l8. 

This  equation,  solved  by  the  usual  method,  gives 
x=4±V— 2,  an  imaginary  expression. 
Hence  we  infer  that  it  is  impossible  to  find  two  numbers 
whose  sum  is  8,  and  product  18.     This  is  obvious  from  the 
Proposition  above  demonstrated,  from  which  it  appears  that 
16  is  the  greatest  product  which  can  be  obtained  by  dividing 
8  into  two  parts,  and  multiplying  them  together. 

3 

(197.)  When  q  is  negative,  and  numerically  equal  to  ^-,  the 
radical  part  of  both  values  of  x  becomes  zero,  and  both  values 
of  a;  reduce  to  —  ~.  The  two  roots  are  then  said  to  be  equal. 

Thus,  in  the  equation 

the  two  roots  are  3  and  3. 

We  say  that  in  this  case  the  equation  has  two  roots,  because 
it  is  the  product  of  the  two  factors,  a;— 3=0,  and  a;— 3=0. 


DISCUSSION  OF  PARTICULAR  PROBLEMS. 

(198.)  In  discussing  particular  problems  which  involve  equa- 
tions of  the  second  degree,  we  meet  with  all  the  different  cases 
which  are  presented  by  equations  of  the  first  degree,  and  some 
peculiarities  besides.  We  may  therefore  have, 

1.  Positive  values  of  x. 

2.  Negative  values. 

3.  Values  of  the  form  of  -r. 

^ 

4.  Values  of  the  form  of  — . 

5.  Values  of  the  form  of  -. 

All  these  different   cases   are   presented   by  Problem   10, 


DISCUSSION    OF    PARTICULAR    PROBLEMS,  IG'J 

page  155,  when  we  make  different  suppositions  upon  the  values 
of  a,  m,  and  n ;  but  we  need  not  dwell  upon  them  here. 

The  peculiarities  exhibited  by  equations  of  the  second  de- 
gree are, 

6.  Double  values  of  x. 

7.  Imaginary  values. 

We  will  consider  the  last  two  cases. 

(199.)  Double  values  of  the  unknown  quantity. 

We  have  seen  that  every  equation  of  the  second  degree  has 
two  roots.  Sometimes  both  of  these  values  are  applicable  to 
the  problem  which  furnishes  the  equation.  Thus,  in  Problem 
20,  page  155,  we  obtain  either  100  or  180  miles  for  the  dis- 
tance between  the  places  C  and  D. 

C  E  D 

I I I 

Let  E  represent  the  situation  of  A  when  B  sets  out  on  his 
journey.  Then,  if  we  suppose  CD  equals  100  miles,  ED  will 
equal  55  miles,  of  which  A  will  travel  30  miles  (being  6  miles 
an  hour  for  5  hours),  and  B  will  travel  25  miles  (being  5  miles 
an  hour  for  5  hours). 

If  we  suppose  CD  equals  180  miles,  ED  will  equal  135  miles, 
of  which  A  will  travel  54  miles  (being  6  miles  an  hour  for  9 
hours),  and  B  will  travel  81  miles  (being  9  miles  an  hour  for 
9  hours). 

This  problem,  therefore,  admits  of  two  positive  answers, 
both  equally  applicable  to  the  question. 

Problem  22,  page  156,  is  of  the  same  kind;  and  another 
will  be  found  on  page  193. 

In  Problem  18,  page  155,  one  of  the  values  of  x  is  positive, 
and  the  other  negative. 

C'  A        C  B 

I  I  I 

Let  the  weakest  magnet  be  placed  at  A,  and  the  strongest 
at  B ;  then  C  will  represent  the  situation  of  a  needle  equally 
attracted  by  both  magnets.  According  to  the  first  value,  the 
distance  AC=8  inches,  and  CB=12.  Now  at  the  distance  of 
R  inches,  the  attraction  of  the  weakest  magnet  will  be  repre- 

4 
sented  by  — ;  and  at  the  distance  of  12  inches,  the  attraction 

o 


170  DISCUSSION    OP    PARTICULAR    PROBLEMS. 

g 

of  the  other  magnet  will  be  represented  by  —»  and  tnese  two 
powers  are  equal ;  foi 

|;<t 

8a     12a* 

But  there  is  another  point,  C',  which  equally  satisfies  the 
conditions  of  the  question ;  and  this  point  is  40  inches  to  the 
left  of  A,  and  therefore  60  inches  to  the  left  of  B ;  for 

±=^- 
40a     60a* 

(200.)  Imaginary  values  of  the  unknown  quantity. 

We  have  seen  that  an  imaginary  root  indicates  an  absurdity 
in  the  proposed  question  which  furnished  the  equation. 

In  several  of  the  preceding  problems,  the  values  of  x  be 
come  imaginary  in  particular  cases. 

When  will  the  values  of  x  in  Problem  6,  page  153,  be  im- 
aginary ? 

Ans.  When  b>a\ 

What  is  the  absurdity  involved  in  this  supposition  ? 

Ans.  It  is  absurd  to  suppose  that  the  product  of  two  num- 
bers can  be  greater  than  the  square  of  half  their  sum. 

When  will  the  values  of  x  in  Problem  11,  page  154,  be  imag- 
inary ? 

Ans.  When  aa>6;  or  (2<z)a>4&. 

What  is  the  absurdity  of  this  supposition  ? 
Ans.  The  square  of  the  sum  of  two  numbers  can  not  be 
greater  than  twice  the  sum  of  their  squares. 

When  will  the  values  of  x  in  Problem  17,  page  155,  be  im- 
aginary ? 

Ans.  When  a*>b;  or  (2a)*>Qb. 

What  is  the  absurdity  of  this  supposition  ? 
Ans.    The  cube  of  the  sum  of  two  numbers  can  not  be 
greater  than  four  times  the  sum  of  their  cubes. 

When  will  the  values  of  x  in  Problem  4,  page  140,  be  im- 
aginary, and  what  is  the  absurdity  of  this  supposition  ? 


SECTION  XIII. 


RATIO  AND  PROPORTION. 

Numbers  may  be  compared  in  two  ways  :  either  by 
means  of  their  difference,  or  by  their  quotient.  We  may  in- 
quire how  much  one  quantity  is  greater  than  another ;  or,  how 
many  times  the  one  contains  the  other.  One  is  called  Arith- 
metical, and  the  other  Geometrical  Ratio. 

The  difference  between  two  numbers  is  called  their  Arith- 
metical Ratio.  Thus,  the  arithmetical  ratio  of  9  to  7  is  9— 7, 
or  2 ;  and  if  a  and  b  designate  two  numbers,  their  arithmetical 
ratio  is  represented  by  a— b. 

Numbers  are  more  generally  compared  by  means  of  quo- 
tients ;  that  is,  by  inquiring  how  many  times  one  number  con- 
tains another.  The  quotient  of  one  number  divided  by  another 
is  called  their  Geometrical  Ratio.  The  term  Ratio,  when  used 
without  any  qualification,  is  always  understood  to  signify  a 
geometrical  ratio,  and  we  shall  confine  our  attention  to  ratios 
of  this  description. 

(202.)  By  the  ratio. of  two  numbers,  then,  we  mean  the  quo- 
tient which  arises  from  dividing  one  of  these  numbers  by  the 
other. 

12 

Thus,  the  ratio  of  12  to  4  is  represented  by  — ,  or  3. 

The  ratio  of  5  to  2  is  -,  or  2.5. 
The  ratio  of  1  to  3  is  -,  or  .333,  &c. 

o 

We  here  perceive  that  the  value  of  a  ratio  can  not  always 
be  expressed  exactly  in  decimals ;  but,  by  taking  a  sufficient 


173  RATIO    AND    PROPORTION. 

number  of  terms,  we  can  approach  as  nearly  as  we  please  to 
the  true  value. 

If  a  and  b  designate  two  numbers,  the  ratio  of  a  to  b  is  the 
quotient  arising  from  dividing  a  by  6,  and  may  be  represented 

by  writing  them  a :  b,  or  7.     The  first  term,  «,  is  called  the 

antecedent  of  the  ratio  ;  the  last  term,  b,  is  called  the  consequent 
of  the  ratio. 

Hence  it  appears  that  the  theory  of  ratios  is  included  in  the 
theory  of  fractions,  and  a  ratio  may  be  considered  as  a  fraction 
whose  numerator  is  the  antecedent,  and  whose  denominator  is  the 
consequent. 

(203.)  When  the  antecedent  of  a  ratio  is  greater  than  the 
consequent,  the  ratio  is  called  a  ratio  of  greater  inequality ;  as, 

5   12 

-,  — .     When  the  antecedent  is  less  than  the  consequent,  it  is 

2   5 

called  a  ratio  of  less  inequality ;  as,  -,  -.     When  the  antece- 

O     y 

dent  and  consequent  are  equal,  it  is  called  a  ratio  of  equality ; 

Q        fi 

as,  o»  o'     I*  is  plain  that  a  ratio  of  equality  may  always  be 

o     o 

represented  by  unity. 

(204.)  When  the  corresponding  terms  of  two  or  more  sim- 
ple ratios  are  multiplied  together,  the  ratios  are  said  to  be 

compounded.     Thus,  the  ratio  of  r,  compounded  with  the  ratio 

„  c  ac 

of  -,  becomes  7-3. 

a  oa 

When  a  ratio  is  compounded  with  itself,  the  result  is  called 

2      4 

a  duplicate  ratio.     Thus,  the  duplicate  ratio  of  -  is  -  ;  and  the 

o     y 

duplicate  ratio  of  7  is  -75. 
b      b 

A  ratio  compounded  of  three  equal  ratios  is  called  a  tripli- 

2       8 
cate  ratio.     Thus,  the  triplicate  ratio  of  -  is  —  ;  and  the  tripli- 

ca.    a3 

cate  ratio  of  7  is  7^. 
o      o 

The  ratio  of  the  square  roots  of  two  quantities  is  called  a 


EATIO    AND    PROPORTION.  173 

4      2 

subduplicate  ratio.     Thus,  the  subduplicate  ratio  of  -  is  - ;  and 

9       3 

the  subduplicate  ratio  of  •=-  is  — j-. 

The  ratio  of  the  cube  roots  of  two  quantities  is  called  a  sub- 

8        2 
triplicate  ratio.     Thus,  the  subtriplicate  ratio  of  —  is  - ;  and 

the  subtriplicate  ratio  of  7  is  -TTT- 

(205.)  If  the  terms  of  a  ratio  are  both  multiplied,  or  both  di- 
vided by  the  same  quantity,  the  value  of  the  ratio  remains  un- 
changed. 

The  ratio  of  a  to  b  is  represented  by  the  fraction  T,  and  the 

value  of  a  fraction  is  not  changed  if  we  multiply  or  divide  both 
numerator  and  denominator  by  the  same  quantity.     Thus, 

a 


n 

a    b 

or  a  :  o=ma  :  mo=-  :  -. 

n   n 

(206.)  Ratios  are  compared  with  each  other  by  reducing 
the  fractions  which  represent  them  to  a  common  denominator. 

In  order  to  ascertain  whether  the  ratio  of  2  to  7  is  greater 
or  less  than  that  of  3  to  8,  we  represent  these  ratios  by  the 

2  3 

fractions  -  and  -,  and  reduce  them  to  a  common  denominator. 

7  o 

They  thus  become 

16       ,  21 


and,  since  the  latter  of  these  is  the  greatest,  we  infer  that  the 
ratio  of  2  to  7  is  less  than  the  ratio  of  3  to  8. 

(207.)  A  ratio  of  greater  inequality  is  diminished,  and  a  ratio 
of  less  inequality  is  increased,  by  adding  the  same  quantity  to 
both  terms. 


174  RATIO    AND    PROPORTION. 

22+1        3 


To  prove  the  proposition  generally,  let  T  represent  any  ra- 

tio, and  let  x  be  added  to  each  of  its  terms.     The  two  ratios 
will  then  be 

a         a+x 


which,  reduced  to  a  common  denominator,  become 

,  .  ,  ,  , 

ao-\-ax     ab-\-bx 

&(6+a5'  b(b+x)' 

Now  if  «>&,  that  is,  if  =-  is  a  ratio  of  greater  inequality,  then, 
since  ax  is  greater  than  bx,  the  first  of  these  fractions  is  great- 
er than  the  second,  and  therefore  r  is  diminished  by  the  addi- 
tion of  the  same  quantity  to  each  of  its  terms. 

But  if  a<&,  that  is,  if  =-  is  a  ratio  of  less  inequality,  then, 
since  ax  is  less  than  bx,  the  first  of  the  above  fractions  is  less 
than  the  second,  and  therefore  7  is  increased  by  the  addition 
of  the  same  quantity  to  each  of  its  terms. 

(208.)  If,  in  a  series  of  ratios,  the  consequent  of  each  is  the 
antecedent  of  the  following  ratio,  then  the  ratio  of  the  first  an- 
tecedent to  the  last  consequent  is  equal  to  that  which  is  corn,' 
pounded  of  all  the  intervening  ratios. 

Let  the  proposed  ratios  be 

a  b  c  d  e 

b'  c'  d'  e'  f 

Compounding  them  by  Art.  204,  we  obtain 

abcde 


which,  being  divided  by  be  de,  reduces  to 

i 


RATIO   AND   PROPORTION.  175 


PROPORTION. 

(209.)  Proportion  is  an  equality  of  ratios. 
Thus,  if  a,  b,  c,  d  are  four  quantities,  such  that  «,  when  di- 
vided by  b,  gives  the  same  quotient  as  c  when  divided  by  d, 
then  a,  b,  c,  d  are  called  proportionals,  and  we  say  that  a  is  to 
b  as  c  is  to  d;  and  this  is  expressed  by  writing  them  thus  : 

a  :  b  : :  c  :  d, 

or  a :  b=c  :  d, 

a    c 

or  i=d- 

So,  also,  3,  4,  9,  12  are  proportionals  ;  that  is, 
3:4::9:12 
3     9 
4=l2* 

In  ordinary  language,  the  terms  ratio  and  proportion  are 
confounded  with  each  other.  Thus,  two  quantities  are  said  to 
be  in  the  proportion  of  3  to  5,  instead  of  the  ratio  of  3  to  5. 
A  ratio  subsists  between  two  quantities,  a  proportion  only  be- 
tween four.  Ratio  is  the  quotient  arising  from  dividing  one 
quantity  by  another  ;  two  equal  ratios  form  a  proportion. 

(210.)  In  the  proportion 

a  :  b  : :  c  :  d, 

a,b,c,d  are  called  the  terms  of  the  proportion.  The  first  and 
last  terms  are  called  the  extremes,  the  second  jand  third  the 
means.  The  first  term  is  called  the  first  antecedent,  the  second 
term  the  first  consequent,  the  third  term  the  second  antecedent, 
and  the  fourth  term  the  second  consequent. 

The  word  term,  when  applied  to  a  proportion,  is  used  in  a 
slightly  different  sense  from  that  v  explained  in  Art.  27.     The 
terms  of  a  proportion  may  be  polynomials.     Thus, 
a+b:c+d::e+f:g+h. 

(211.)  When  the  second  and  third  terms  of  a  proportion  are 
identical,  this  quantity  is  called  a  mean  proportional  between 
the  other  two.  Thus,  if  we  have  three  quantities,  a,  b,  c,  such 
that 

a  :  b  : :  b  :  c, 


176  RATIO    AND   PROPORTION. 

then  b  is  called  a  mean  proportional  between  a  and  c,  and  c  is 
called  a  third  proportional  to  a  and  b. 

If,  in  a  series  of  proportional  magnitudes,  each  consequent  is 
identical  with  the  next  antecedent,  these  quantities  are  said  to 
be  in  continued  proportion.  Thus,  if  we  have  a,  ft,  c,  d,  e,  f 
such  that 

a  :  b  ::  b  :  c  ::  c  :  d  ::  d  :  e  ::  e  :/, 

a    b    c     d     e 
or  -=-=-=-=—, 

the  quantities  a,  b,  c,  d,  e,  f  are  in  continued  proportion. 

(212.)  If  four  quantities  are  proportional,  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means. 

Let  a  :  b  : :  c  :  d. 

Then  will  ad=bc. 

For,  since  the  four  quantities  are  proportional, 

••tt^'i' 
a    c 

b     d' 

Multiplying  each  of  these  equals  by  bd,  the  expression  be- 
comes 

abd_bcd 

or  ad=bc. 

Thus,  if  3  :  4  : :  9  :  12, 

then  3X12=4X9. 

(213.)  Conversely,  if  the  product  of  two  quantities  is  equal 
to  the  produQt  of  two  others,  the  first  two  quantities  may  be 
made  the  extremes,  and  the  other  two  the  means  of  a  proportion. 

Let  ad=bc. 

Then  will  a  :  b  : :  c  :  d. 

For,  since  ad=bc, 

dividing  each  of  these  equals  by  bd,  the  expression  becomes 
a_c        c  _a 

that  is,  a  :  b  : :  c  :  d,  or  c  :  d  : :  a  :  b. 

Thus,  if  3X12=4X9, 

then  3  :  4  : :  9  :  12, 

or  9  :  12  :  :  3  :  4. 


RATIO    AND    PROPORTION.  177 

(214.)  The  preceding  proposition  is  called  the  test  of  propor- 
tions, and  any  change  may  be  made  in  the  form  of  a  propor- 
tion which  is  consistent  with  the  application  of  this  test.  In 
order,  then,  to  decide  whether  four  quantities  are  proportional, 
we  must  compare  the  product  of  the  extremes  with  the  product 
of  the  means. 

Thus,  to  determine  whether  5,  6,  7,  8  are  proportional,  we 
multiply  5  by  8,  and  obtain  40.  Multiplying  6  by  7,  we  ob- 
tain 42.  As  these  two  products  are  not  equal,  we  conclude 
that  the  numbers  5,  6,  7,  8  are  not  proportional. 

Again,  take  the  numbers  5,  6,  10,  12.  The  product  of  5  by 
12  is  60,  and  the  product  of  6  by  10  is  also  60.  Hence  these 
numbers  are  proportional ;  that  is, 

5:6::  10:  12. 

(215.)  If  three  quantities  are  in  continued  proportion,  the 
product  of  the  extremes  is  equal  to  the  square  of  the  mean. 

If  a:b::b:c. 

Then,  by  Art.  212,        ac=bb,  which  is  equal  to  b'\ 

Conversely,  if  the  product  of  two  quantities  is  equal  to  the 
square  of  a  third,  the  last  quantity  is  a  mean  proportional  be- 
tween the  other  two. 

Thus,  let  ac=b\ 

Dividing  these  equals  by  be,  we  obtain 

a_b 
ft"? 
or  a  :b  : :  b  :  c. 

Thus,  if  4  :  6  :  :  6  :  9, 

then  4X9:=62. 

And  conversely,  if        4X9=62, 
then  6  is  a  mean  proportional  between  4  and  9. 

EXAMPLES. 

1.  Given  the  first  three  terms  of  a  proportion,  24,  15,  and 
40,  to  find  the  fourth  term. 

2.  Given  the  first  three  terms  of  a  proportion,  3«fe3,  4a26% 
and  9asb,  to  find  the  fourth  term. 

3.  Given  the  last  three  terms  of  a  proportion,  4<z366,  3a36% 
and  2a*b,  to  find  the  first  term. 


178  RATIO    AND    PROPORTION. 

4.  Given  the  first,  second,  and  fourth  terms  of  a  proportion, 
5y4,  7x*y*,  and  21x*y,  to  find  the  third  term. 

5.  Given  the  first,  third,  and  fourth  terms  of  a  proportion, 
a+b,  a*—b*,  and  (a— 6)3,  to  find  the  second  term. 

(216.)  Ratios  that  are  equal  to  the  same  ratio  are  equal  to 
each  other. 

(  then  will  a:b::c:d. 

and  c  :  d : :  x  :  y,  ) 

For,  since  a:b::x:y9 

ax 

we  have  r— ~» 

b    y 

And  since  c:d::x:yt 

c    x 

we  have  -3=-. 

d    y 

a    c 

Therefore,  r=3» 

o     a 

and  hence  a  :  b  : :  c  :  d. 

(217.)  If  four  quantities  are  proportional,  they  will  be  pro- 
portional by  alternation;  that  is,  the  first  will  have  the  same  ratio 
to  the  third  that  the  second  has  to  the  fourth. 

Let  a  :  b  : :  c  :  d9 

then  will  a  :  c  : :  b :  d. 

For  since  a  :  b  : :  c  :  df 

by  Art.  212,  ad=bc9 

and  since  ad=bc9 

by  Art.  213,  a:e::b:d. 

(218.)  If  four  quantities  are  proportional,  they  will  be  pro- 
portional by  inversion ;  that  is,  the  second  will  have  to  the  first 
the  same  ratio  that  the  fourth  has  to  the  third. 

Let  a  :  b  : :  c  :  d; 

then  will  b  :  a : :  d :  c. 

T*  t  J 

For  since  a  :  b  : :  c  :  d9 

by  Art.  212,  ad=bc9 

or  be— ad. 

Therefore,  by  Art.  213,    b:a::d:c. 

(219.)  If  four  quantities  are  proportional,  they  will  be  pro- 


RATIO    AND    PROPORTION.  I'll) 

portional  by  composition  ;  that  is,  the  sum  of  the  first  and  sec- 
ond will  have  to  the  second  the  same  ratio  that  the  sum  of  the 
third  and  fourth  has  to  the  fourth. 

Let  a  :  b  : :  c  :  d  ; 

then  will  a+b  :b  ::  c+d :  d. 

For  since  a  :  b  : :  c  :  d, 

a    c 

we  have  T=~;» 

b    a 

Add  unity  to  each  of  these  equals,  and  we  have 

a          c  a+b    c+d 

_+1=_+1,  or  __::__,. 

that  is,  a+b:b::c+d:d. 

(220.)  If  four  quantities  are  proportional,  they  will  be  pro- 
portional by  division;  that  is,  the  difference  of  the  first  and  sec- 
ond will  have  to  the  second  the  same  ratio  that  the  difference  of 
the  third  and  fourth  has  to  the  fourth. 

Let  a  :  b  : :  c  :  d  ; 

then  will  a—b  :b::  c—d :  d. 

For  since  a  :  b  : :  c  :  d, 

a    c 

we  have  •=-= > 

b     a 

Subtract  unity  from  each  of  these  equals,  and  we  have 

a          c  a—b    c—d 

--l=-_l,or— =^-; 

that  is,  a—b  :b  ::  c—d :  d. 

(221.)  If  four  quantities  are  proportional,  they  will  be  pro- 
portional by  conversion ;  that  is,  the  first  will  have  to  the  dif- 
ference of  the  first  and  second  the  same  ratio  that  the  third  has 
to  the  difference  of  the  third  and  fourth. 

Let  a:b::c:d; 

then  will  a  :  a—b  :  :c  :  c—d. 

For  since  a  :  b  : :  c  :  d9 

by  inversion,  b  :  a  : :  d :  c  ; 

b  d 

whence  -=-. 

a    c 

Subtract  each  of  these  equals  from  unity,  and  we  have 


180  RATIO    AND    PROPORTION'. 

b          d        a—b    c—d 
1  —  =1  —  ,  or  -  =  -  ; 
a          c  a          c 

that  is,  a—b  :  a  ::  c—d  :  c, 

or  inversely,  a  :  a—b  ::  c  :  c—d. 

(222.)  If  four  quantities  are  proportional,  the  sum  of  the  first 
and  second  will  have  to  their  difference  the  same  ratio  that  the 
sum  of  the  third  and  fourth  has  to  their  difference. 


hh 


Let 

a:b: 

c-.d; 

then  will                 a+b  :  a—b  : 

c+d  :  c-d. 

For  since 

a:b: 

c:d, 

by  composition, 

a+b  :  b  : 

c+d  :  d, 

and  by  alternation, 

a+b  :  c+ 

i::b:d. 

Also,  since 

a:b: 

c:d, 

by  division, 

a—b  :  b  : 

c—d  :  d. 

and  by  alternation, 

a—b  :  c—d  ::  b  :  d. 

Hence,  by  equality  of  ratios, 

a+b  :  a—b  :  c+d  :  c—d. 

(223.)  If  four  quantities  are  proportional,  like  powers  or  roots 
of  these  quantities  will  also  be  proportional. 

Let  a  :  b  : :  c  :  d  ; 

then  will  an  :  bn  : :  cn  :  dn. 

For  since  a  :  b  : :  c  :  d, 

a    c 

we  have  T= > 

o    a 

Raising  each  of  these  equals  to  the  rath  power,  we  obtain 

——— 
lr~I«; 

that  is,  an:bn::cn:  d\ 

where  n  may  be  either  a  whole  number  or  a  fraction. 

(224.)  If  there  is  any  number  of  proportional  quantities  all 
having  the  same  ratio,  the  first  will  have  to  the  second  the  same, 
ratio  that  the  sum  of  all  the  antecedents  has  to  the  sum  of  all  the 
consequents. 

Let  a,  &,  c,  d,  e,  f  be  any  number  of  proportional  quantities 
such  that 

a  :  b  ::  c  :  d ::  e  :/, 

then  will  a  :  b  : :  a+c+e  :  b+d+f. 


RATIO    AND    PROPORTION.  181 

For  since  a  :  b  : :  c  :  d, 

we  have  ad=bc; 

and  since  a  :  b  : :  e  :/, 

we  have  af=be. 

To  these  equals  add         ab=ba, 
and  we  obtain         a(b+d+f)=b(a+c+e). 

Hence,  by  Art.  213,  a  :  b  : :  a*\-c+e  :  b+d+f. 

(225.)  If  three  quantities  are  in  continued  proportion,  the 
first  will  have  to  the  third  the  duplicate  ratio  of  that  which  it 
has  to  the  second. 

Let  a  :  b  : :  b  :  c. 

Then  a  :  c  : :  aa  :  b*. 

For  since  a  :  b  : :  b  :  c, 

by  Art.  212,  ac=b*. 

Multiplying  each  of  these  equals  by  a,  we  obtain 

aac=a&3; 
that  is,  a*Xc=aXb*. 

Resolving  this  equation  into  a  proportion  by  Art.  213,  we 
have 

a  :  c  : :  a*  :  b\ 

(226.)  If  four  quantities  are  in  continued  proportion,  the  first 
will  have  to  the  fourth  the  triplicate  ratio  of  that  which  it  has  to 
the  second. 

Let  a,  bj  c,  d  be  four  quantities  in  continued  proportion,  so 
that 

a  :b  ::  b  :  c  ::  c  :  d, 
then  will  a  :  d  : :  a9  :  b9. 

For  since  a  :  b  : :  c  :  d, 

we  have  ad=bc; 

and  since  a  :  b  : :  b  :  c, 

we  have  ac=b*. 

Multiplying  these  equals  by  ab,  we  obtain 

a\bdc)=b\abc\ 
or  a*Xd=b*Xa. 

Hence,  by  Art.  213,     a  :  d  - :  a9  :  b3. 


182  RATIO    AND   PROPORTION. 

(227.)  If  there  are  two  sets  of  proportional  quantities,  the 
products  of  the  corresponding  terms  will  be  proportional. 

Let  a  :  b  : :  c  :  d, 

and  e:fi:g:h. 

Then  will  ae:bf::cg:  dh. 

For,  since  a  :  b  : :  c  :  d, 

by  Art.  212,  ad=bc. 

And  since  e  :f  ::  g  :  h, 

by  Art.  212,  eh=fg. 

Multiplying  these  equals  together,  we  have     . 

aeXdh=bfXcg. 
Hence,  by  Art.  213,      ae  :  bf : :  eg  :  dh. 

(228.)  Three  quantities  are  said  to  be  in  harmonical  propor- 
tion when  the  first  is  to  the  third  as  the  difference  between  the 
first  and  second  is  to  the  difference  between  the  second  and  third. 

Thus,  2,  3,  6  are  in  harmonical  proportion  ;  for 
2:  6::  3-2  :  6-3. 

Let  a,  6,  c  be  in  harmonical  proportion ;  then 

*v  i.     i. 

a  :  c  : :  a—b  :  b—c. 

Multiplying  the  extremes  and  means,  and  reducing,  we  have 

c—"z. r- 

2a— b 

Hence,  to  find  a  third  harmonical  proportional  to  two  quan- 
tities, divide  the  product  of  the  first  and  second  by  twice  the 
first  diminished  by  the  second. 

Ex.  1.  Find  a  third  harmonical  proportional  to  3  and  5. 

Ex.  2.  Find  a  third  harmonical  proportional  to  5  and  8. 

(229.)  Four  quantities  are  said  to  be  in  harmonical  propor- 
tion when  the  first  is  to  the  fourth  as  the  difference  between  the 
first  and  second  is  to  the  difference  between  the  third  and  fourth. 

Thus,  2,  3,  4,  8  are  in  harmonical  proportion ;  for 
2:8::  3-2  :  8-4. 

Let  «,  &,  c,  d  be  in  harmonical  proportion ;  then 
aid:'  a—b  :  c—d. 

Multiplying  the  extremes  and  means,  and  reducing,  we 
have 


RATIO    AND   PROPORTION.  183 

> 

.'. 
,.i   cJl. 

Hence,  to  find  a  fourth  harmonical  proportional  to  three 
quantities,  divide  the  product  of  the  first  and  third  by  twice 
the  first  diminished  by  the  second. 

Ex.  1.  Find  a  fourth  harmonical  proportional  to  4,  5,  and  6. 

Ex.  2.  Find  a  fourth  harmonical  proportional  to  5, 8,  and  10. 

(230.)  Proportions  are  often  expressed  in  an  abridged  form. 
Thus,  if  A  and  B  represent  two  sums  of  money  put  out  for 
one  year  at  the  same  rate  of  interest,  then 

A  :  B  : :  interest  of  A  :  interest  of  B. 

This  is  briefly  expressed  by  saying  that  the  interest  varies 
as  the  principal.  A  peculiar  character  QD  is  used  to  denote 
this  relation.  Thus,  we  write  .,.  ,  v 

the  interest  QD  the  principal. 

One  quantity  varies  directly  as  another,  when  both  increase 
or  diminish  together  in  the  same  ratio.  Thus,  in  the  above 
example,  A  varies  directly  as  the  interest  of  A.  In  such  a  case 
either  quantity  is  equal  to  the  other  multiplied  by  some  con- 
stant number.  Thus,  if  the  interest  varies  as  the  principal, 
then  the  interest  equals  the  principal  multiplied  by  a  constant 
quantity,  which  is  the  rate  of  interest. 

If  A  QD  B,  then  A=wB. 

If  the  space  (S)  described  by  a  falling  body  varies  as  the 
square  of  the  time  (T),  then 


m  representing  some  constant  quantity. 

(231.)  One  quantity  may  vary  directly  as  the  product  of 
several  others.  Thus,  if  a  body  moves  with  uniform  velocity, 
the  space  described  is  measured  by  the  product  of  the  time 
by  the  velocity.  If  we  put  S  to  represent  the  space  described, 
T  the  time  of  motion,  and  V  the  uniform  velocity,  then  we 
shall  have 

SooTxV. 

Also  the  area  of  a  rectangle  varies  as  the  product  of  its 
length  and  breadth. 

The  weight  of  a  stick  of  timber  varies  as  its  length  X  its 
breadth  X  its  depth  X  its  density. 

9 


184  RATIO    AND    PROPORTION. 

If  the  density  is  given,  then  the  weight  varies  as  the  length 
X  the  breadth  X  the  depth. 

If  the  depth  also  is  given,  then  the  weight  varies  as  the  length 
X  the  breadth. 

If  the  breadth  is  given,  then  the  weight  varies  as  the  length. 

Finally,  if  the  length  also  is  given,  then  the  weight  is  equal 
to  a  constant  quantity. 

(232.)  One  quantity  varies  inversely  as  another  when  one 
increases  in  the  same  ratio  that  the  other  diminishes.  Thus, 
the  altitude  of  a  triangle  whose  area  is  given,  varies  inversely 
as  its  base. 

If  the  product  of  two  quantities  is  constant,  then  one  varies 

inversely.  as  the  other. 
j 

In  uniform  motion,  the  space  is  measured  by  the  product  of 
the  time  by  the  velocity  ;  that  is, 


Whence  T=. 

If  the  space  be  supposed  to  remain  constant,  then 


that  is,  the  time  required  to  travel  a  given  distance  varies  in- 
versely as  the  velocity.  Suppose  the  distance  is  360  miles  : 
then, 

if  the  velocity  is  12  miles  per  hour,  the  time  will  be  30  hours  ; 
"  20  "  "  18      " 

"  24  "  "  15      " 

that  is,  if  the  velocity  is  doubled,  the  time  is  halved.  The  one 
varies  inversely  as  the  other. 

Conversely,  if  one  quantity  varies  inversely  as  another,  the 

product  of  the  two  quantities  is  constant. 

n  .     3fltrr  9*i;  ', 

Thus,  if  T  OD  i 

then  the  space  (S)  is  a  constant  quantity. 

(233.)  One  quantity  may  vary  directly  as  a  second,  and  in- 
versely as  a  third.  Thus,  according  to  the  Newtonian  law  of 
gravitation,  the  attraction  (G)  of  any  heavenly  body  varies 

' 


RATIO    AND   PROPORTION.  185 

directly  as  the  quantity  of  matter  (Q),  and  inversely  as  the 
square  of  the  distance  (D). 

That  is,  G  oo  ^. 

(234.)  Application  of  the  preceding  principles. 

Ex.  1.  Given  x+yix:-.  5:3,  j 

|  to  find  the  values  of  a:  and  y. 

Since  x+y  :  x  :  :  5  :  3. 

By  division,  Art.  220,    y  :  x  :  :  2  :  3. 

Q/w 

Therefore,  3y=2#,  and  y=-^-. 

o 

Substituting  this  value  of  y  in  the  second  equation,  we  obtain 

|U 

Therefore,  z 

and  y 

jBa:.  2.  Given  x+y  :  x  —y  :  :  3  :  1,  /  to  find  the  values  of  x 

x9—  y3=56,      S      and  y. 
From  the  first  equation,  by  Art.  222,  we  obtain 

2x  :  2y  :  :  4  :  2  ; 

whence,  #  :  y  :  :  2  :  1, 

and  x=2y. 

Substituting  this  value  of  x  in  the  second  equation,  we  ob- 
tain 


Ex.  3.  Given  x+y  :  x—  y  :  :  64  :  1,  j  to  find  the  values  of  a: 

xy=  63,        )      and  y. 
By  Art.  223,  x+y  :  x—y  :  :  8  :  1. 

By  Art.  222,  2x  :  2y  :  :  9  :  7  ; 

whence  a;  :  y  :  :  9  :  7. 


Therefore,  *=T. 

Substituting  this  value  of  x  in  the  second  equation,  we  ob- 
tain 


186  RATIO    AND    PROPORTION. 


Ex.  4.  Given  x*— y* :  x— y  : :  61  :  1,  |  to  find  the  values  of 

xy=320,     )      x  and  y. 

\J?.  f^  •      A  A»r4rjT' 

Since         a;3-v3 :  x*—3x*v+3xy*— y9 : :  61  :  1. 


By  division,  Art.  220,  3xyx(x—y)  :  x—y  : :  60  :  1. 

2 

60:  1, 


1:1. 


Hence  960  :  x—y  : 

? 

and  16  :  x—y  : 

Therefore,  x—y  =±4. 

Also,  since 

And 

By  addition,  ; 

Extracting  the  root,     x+y=±36. 

Hence  x=±20,  or  ±16, 

y=±16,  or  ±20. 

Ex.  5.  Given  x3— y3  :  ccay— xya  : :  7  :  2,  >  to  find  the  values 

x+y— 6,        y      ofxandy. 
Ans.  x=4,  or  2  ;  y=2,  or  4. 

Ex.  6.  Given  Vy      —  Va—x=  Vy—x,  )  to  find  the 

Vy—x+  Va—x  :  Va—x  : :  5  :  2,  )    values  of 

x  and  y. 

4#          5a 
4»S.  X=-;y=-. 


Ex.  7.  Given  x+  <Jx  :  x—  <Jx  : :  S^/x+Q  :  2,/x,  to  find  the 
values  of  x. 

Ans.  x=9,  or  4. 

Ex.  8.  What  number  is  that  to  which,  if  1,  5,  and  13  be  sev- 
erally added,  the  first  sum  shall  be  to  the  second  as  the  second 
to  the  third  ? 

Ans.  3. 

Ex.  9.  What  number  is  that  to  which,  if  a,  b,  and  c  be  sev- 
erally added,  the  first  sum  shall  be  to  the  second  as  the  second 
to  the  third  ? 

b*-ac 


Ans. 


-  .      . 
a—  2b+c 


RATIO    AND    PROPORTION.  187 

Ex.  10.  What  two  numbers  are  those  whose  difference,  sum, 
and  product  are  as  the  numbers  2,  3,  and  5  respectively  ? 

Ans.  2  and  10. 

Ex.  11.  What  two  numbers  are  those  whose  difference,  sum, 
and  product  are  as  the  numbers  m,  n,  and  p  ? 

Ans.  —  ^—  ,  and  —  —  . 
n+m  n—m 

Ex.  12.  Find  two  numbers,  the  greater  of  which  shall  be  to 
the  less  as  their  sum  to  42,  and  as  their  difference  to  6. 

Ans.  32  and  24. 

Ex.  13.  Find  two  numbers,  the  greater  of  which  shall  be  to 
the  less  as  their  sum  to  a,  and  their  difference  to  b. 

(a+b)*  a+b 


Ex.  14.  There  are  two  numbers  which  are  in  the  ratio  of  3 
to  2,  the  difference  of  whose  fourth  powers  is  to  the  sum  of 
their  cubes  as  26  to  7.  Required  the  numbers. 

Ans.  6  and  4. 

Ex.  15.  What  two  numbers  are  in  the  ratio  of  m  to  n,  the 
difference  of  whose  fourth  powers  is  to  the  sum  of  their  cubes 
as  p  to  q  ? 

mp    ms+7i3        .  np    m*+n* 

Ans.  —  X—  i  -  1,  and  —  X—  -  :. 

q      m  —n  q     m—n* 


has  $'.£  ''^gdoujft  •  arf  j  as  am  :  J:>tfhQ| 

.';i?k 

9tiiL  saotiw  daod^irur  tiaiiffrwa  owj  Ifiii  W  .  U  .^ 


SECTION  XIV. 

ikda  lioidw  lo  i9l«a'j    aiUsxadaiuu  <»w)  btiil 


P  R  O  G  R  E  S  S  I  O  N  S.  , 

ARITHMETICAL  PROGRESSION. 

(235.)  -Aw  Arithmetical  Progression  is  a  series  of  quantities 
which  increase  or  decrease  by  the  continued  addition  or  subtrac- 
tion of  the  same  quantity. 

rp,  .,  , 

Thus,  the  numbers 

ill       .*    O)  i)V.  ».t\   a^t.lJ;')  liO«11 

1,3,  5,7,9,  11,  &c., 

which  are  obtained  by  the  addition  of  2  to  each  successive 
term,  form  what  is  called  an  increasing  Arithmetical  Progres- 
sion ;  and  the  numbers 

20,  17;  14,  11,8,  5,  &c., 

which  are  obtained  by  the  subtraction  of  3  from  each  success- 
ive term,  form  what  is  called  a  decreasing  Arithmetical  Pro- 
gression. 

(236.)  To  find  the  last  term  of  an  Arithmetical  Progression. 
If  a  represent  the  first  term  of  an  arithmetical  progression, 
and  d  the  common  difference,  the  successive  terms  of  an  in- 
creasing series  will  be 

a,  a+d,  a+2d,  a+3d,  a+4d,  &c. 
The  successive  terms  of  a  decreasing  series  will  be 

a,  a—d,  a—2d,  a—  3d,  a—4d,  &c. 

Since  the  coefficient  of  d  in  the  second  term  is  1,  in  the  third 
term  2,  in  the  fourth  term  3,  and  so  on,  the  nth  term  of  the 
series  will  be 

a±(n-l)d, 

which  may  be  called  the  last  term  when  the  number  of  terms 
•s  n.  Hence, 


PROGRESSIONS.  189 

The  last  term  of  an  arithmetical  progression  is  equal  to  the 
first,  ±  the  product  of  the  common  difference  into  the  number  of 
terms  less  one. 

In  what  follows  we  shall  consider  the  progression  an  increas- 
ing one,  since  all  the  results  which  we  obtain  can  be  immediate- 
ly applied  to  a  decreasing  series  by  changing  the  sign  of  d. 

If  we  put  /  to  represent  the  last  term  of  the  series,  we  shall 
accordingly  have 

l=a+(n-l)d. 

This  equation  contains  four  variable  quantities,  any  one  of 
which  may  be  computed  when  the  other  three  are  known. 

(237.)   To  find  the  sum  of  n  terms  of  the  series. 

Take  any  series,  and  under  it  set  the  same  terms  in  an  in- 
verted order,  thus : 

Let  the  series  be  1,    3,    5,    7,    9,  11,  13,  15, 

the  same  series  inverted  is  15,  13,  11,    9,    7,    5,    3,    1. 

The  sums  are,  16,  16,  16,  16,  16,  16,  16,  16. 

The  sums  of  the  two  series  must  be  double  the  sum  of  a  sin- 
gle series,  and  is  equal  to  the  sum  of  the  extremes  repeated  as 
many  times  as  there  are  terms. 

In  order  to  generalize  this  method,  let  S  represent  the  sum 
of  the  series, 

Then  S=a+a+d+a+2d+a+3d+ +/. 

If  we  write  the  same  series  in  an  inverted  order,  thus : 


S=l+l-d+l-2d+l-3d+  ............  + 

and  add  the  two  series  together,  term  by  term,  we  obtain 


Represent  the  number  of  terms  in  the  series  by  n  ;  then 
2S=n(l+a). 

n(l+a) 
Hence  S=—  *—•  —  '. 

Therefore, 

The  sum  of  an  Arithmetical  Progression  is  equal  to  half  the, 
sum  of  the  two  extremes,  multiplied  by  the  number  of  terms. 

It  also  appears  from  the  above,  that  the  sum  of  the  extremes 
is  equal  to  the  sum  of  any  other  two  terms  equally  distant  from 
the 


190 


PROGRESSIONS. 


(238.)  The  two  fundamental  equations 
l=a+(n—l)d, 

«     *+* 
~~      ' 


contain  five  variable  quantities, 

a,  /,  d,  n,  S, 

of  which  any  three  being  given,  the  other  two  may  be  found. 
Accordingly,  20  different  cases  may  arise,  all  of  which  are 
solved  by  combining  the  formulae  above  given.  These  cases 
are  exhibited  in  the  following  table,  and  should  be  verified  by 
the  student  : 


No. 

Given. 

Required. 

Formula. 

1 

2 
3 

4 

0,  d,  n 
a,d,S 

a,n,S 
d,n,S 

' 

/=«+(»-  i)rf, 

2S 

/=--a, 

f_S  |  (i.-l)df 

5 
6 

7 

8 

a,d,n 
a,d,l 

a,n,l 
d,n,l 

S 

S=4»{2a+(n-l)<fj, 
I+a    ?_«• 

'   2     '    2d   ' 

9 
10 
11 
12 

a,  n^  I 
a,  w,  S 
a,  /,  S 

71,  /,    S 

d 

J 

n-V 
2S-2an 

71(71-1)' 

2n/-2S 

13 
14 
15 
16 

d,n,l 

dj,  S 

71,  /,    S 

a 

_S     (n-l)d 
~n          2      ' 

Is 

a—  /. 

71 

PROGRESSIONS. 


191 


No. 

Given. 

Required. 

Formulae. 

17 

18 
19 
20 

a,  d,  I 
a,d,S 
a,  I,  S 
d,l,  S 

n 

n    '-"+1 

n       d    Ilf 

±  -v/(2«-d)a+8rfS-2a+d 

2rf 
2S 
n=/+? 

2/+rf±V(2/+rf)a-8dS 

71                       2d 

EXAMPLES. 

(239.)  j&c.  1.  Required  the  sum  of  60  terms  of  an  arithmetical 
progression  whose  first  term  is  5,  and  common  difference  10. 

Ans.  18000. 

This  example  affords  an  application  of  Formula  5. 
Ex.  2.  Required  the  number  of  terms  of  a  progression  whose 
sum  is  442,  whose  first  term  is  2,  and  common  difference  3. 

Ans.  17. 

This  example  is  solved  by  Formula  18. 
Ex.  3.  Required  the  first  term  of  a  progression  whose  sum 
is  99,  whose  last  term  is  19,  and  common  difference  2. 

Ans.  3. 

Ex.  4.  The  sum  of  a  progression  is  1455,  the  first  term  5,  and 
the  last  term  92.  What  is  the  common  difference  ? 

Ans.  3. 

Ex.  5.  A  body  falls  16  feet  during  the  first  second,  and  in 
each  succeeding  second  32  feet  more  than  in  the  one  imme- 
diately preceding.  If  it  continue  falling  for  20  seconds,  how 
many  feet  will  it  pass  over  in  the  last  second,  and  how  many 
in  the  whole  time  ? 

Ans.  624  feet  in  the  last  second,  and  6400  feet  in  the  whole 
time. 

Ex.  6.  Required  the  sum  of  101  terms  of  the  series 
1,  3,  5,  7,  9,  &c. 


Ex.  7.  Find  the  nth  term  of  the  series 
1,  3,  5,  7,  9,  &c. 


9* 


Ans.  10201. 


Ans.  2n—l ; 


102  PROGRESSIONS. 

that  is,  the  last  term  of  this  series  is  one  kss  than  twice  the  num- 
ber of  terms.  Zz^a^ 

Ex.  8.  Find  the  sum  of  n  terms  of  the  series 
1,  3,  5,  7,  9,  &c. 

Ans.  n9; 

that  is,  the  sum  of  the  terms  of  this  series  is  equal  to  the  square 
of  the  number  of  terms. 

Thus,  1+3  =  4=2". 

1+3+5  =  9=3'. 

1+3+5+7       =  16=4a. 

1+3+5+7+9=25=5'. 

Ex.  9.  Find  the  sum  of  the  natural  series  of  numbers 

1,  2,  3,  4,  5,  &c., 
up  to  n  terms. 


Ex.  10.  Find  the  sum  of  the  even  numbers 

2,  4,  6,  8,  &c., 
up  to  n  terms. 

Ans.  n(n+l). 

Ex.  11.  One  hundred  stones  being  placed  on  the  ground  in 
a  straight  line,  at  the  distance  of  two  yards  from  each  other, 
how  far  will  a  person  travel  who  shall  bring  them  one  by  one 
to  a  basket  which  is  placed  two  yards  from  the  first  stone  ? 

Ans.  20200  yards. 

El:    uH&  J>1/-  .  .          J   . 

Ex.  12.  Find  m  arithmetical  means  between  two  given  num- 
bers. 

In  order  to  solve  this  problem,  we  must  first  find  the  com- 
mon difference.  The  whole  number  of  terms  consists  of  the 
two  extremes  and  all  the  intermediate  terms.  If,  then,  m  rep- 
resent the  number  of  means,  m+2  will  be  the  whole  number 
of  terms,,,;,  w  3r 

Substituting  m+2  for  n,  in  Formula  9,  page  190,  we  have 

2*11'  .••»»?!-     I—  a 

d=  —  r-r=  the  common  difference, 
m+1 

whence  .the  required  means  are  easily  obtained  by  addition. 
Ex.  13.  Find  6  arithmetical  means  between  1  and  50. 


PROGRESSIONS.  193 

Ex.  14.  Find  three  numbers  in  arithmetical  progression,  the 
sum  of  whose  squares  shall  be  1232,  and  the  square  of  the 
mean  greater  than  the  product  of  the  two  extremes  by  16. 

Ans.  16,  20,  and  24. 

In  examples  of  this  kind,  it  is  generally  best  to  represent  the 
series  in  such  a  manner  that  the  common  difference  may  dis- 
appear in  taking  the  sum  of  the  terms.     Thus  a  progression 
of  three  terms  may  be  represented  by 
a— d,  a,  a-\-d; 
one  of  four  terms  by  a— 3d,  a—d,  a+d,  a+3d,  &c. 

Ex.  15.  Find  three  numbers  in  arithmetical  progression,  the 
sum  of  whose  squares  shall  be  «,  and  the  square  of  the  mean 
greater  than  the  product  of  the  two  extremes  by  b. 

/a-2b  /a-2b  /a-2b 

.   y  — 3 Vb;  \/~3~;  and  V  "3"+  Vb' 

Ex.  16.  Find  four  numbers  in  arithmetical  progression 
whose  sum  is  28,  and  continued  product  585. 

Ans.  1,  5,  9,  13. 

Ex.  17.  A  sets  out  for  a  certain  place,  and  travels  1  mile 
the  first  day,  2  the  second,  3  the  third,  and  so  on.  In  five  days 
afterward  B  sets  out,  and  travels  12  miles  a  day.  How  long 
will  A  travel  before  he  is  overtaken  by  B  ? 

Ans.  8  or  15  days. 

This  is  another  example  of  an  equation  of  the  second  de- 
gree, in  which  the  two  roots  are  both  positive.  The  following 
diagram  exhibits  the  daily  progress  of  each  traveler.  The  di- 
visions above  the  horizontal  line  represent  the  distances  trav- 
eled each  day  by  A ;  those  below  the  line  the  distances  trav- 
eled by  B. 


A. 
B. 

1234 

II  1   1 

5 

1 

6    7 
1      1 

8 
1 

9 

1 

10 

1 

11 

1 

12 

1 

13 

1 

14          1 

1 

1 
1 

1 
2 

3 

4 

1 
5 

6 

1 
7 

-8 

1 
9         1 

It  is  readily  seen  from  the  figure  that  A  is  in  advance  of  B 
until  the  end  of  his  8th  day,  when  B  overtakes  and  passes  him. 
After  the  12th  day,  A  gains  upon  B,  and  passes  him  on  the 
15th  day,  after  which  he  is  continually  gaining  upon  B,  and 
could  not  be  again  overtaken. 

Ex.  18.  A  goes  1  mile  the  first  day,  2  the  second,  and  so  on. 

N 


194  PROGRESSIONS. 

B  starts  a  days  later,  and  travels  b  miles  per  day.  How  long 
will  A  travel  before  he  is  overtaken  by  B  ? 

2b-I±V(2b-iy-8ab  , 
Ans.  *~ days. 

In  what  case  would  B  never  overtake  A  ? 

Ans.  When  «>M-J£ 

'    9c«  ""      h- 

For  instance,  in  the  preceding  example,  if  B  had  started  one 
day  later,  he  could  never  have  overtaken  A. 

Ex.  19.  A  traveler  set  out  from  a  certain  place  and  went  1 
mile  the  first  day,  3  the  second,  5  the  third,  and  so  on.  After 
he  had  been  gone  three  days,-  a  second  traveler  sets  out,  and 
goes  12  miles  the  first  day,  13  the  second,  and  so  on.  In  how 
many  days  will  the  second  overtake  the  first  ? 

Ans.  In  2  or  9  days. 

Let  the  student  illustrate  this  example  by  a  diagram  like  the 
preceding. 

GEOMETRICAL  PROGRESSION. 

(240.)  A  Geometrical  Progression  is  a  series  of  quantities, 
each  of  which  is  equal  to  the  product  of  that  which  precedes  it  by 
a  constant  number.  >  yrj  < 

Thus,  the  series 

2,  4,  8,  16,  32,  &c., 

and  81,  27,  9,  3,  &c., 

are  geometrical  progressions.  In  the  former,  each  number  is 
derived  from  the  preceding  by  multiplying  it  by  2,  and  the 
series  forms  an  increasing  geometrical  progression.  In  the 
latter,  each  number  is  derived  from  the  preceding  by  multiply- 
ing it  by  J,  and  the  series  forms  a  decreasing  geometrical  pro- 
gression. 

In  each  of  these  cases,  the  common  multiplier  is  called  the 
common  ratio. 

(241.)   To  find  the  last  term  of  a  geometrical  progression. 

Let  a  represent  the  first  term  of  the  progression,  and  r  the 
common  ratio ;  then  the  successive  terms  of  the  series  will  be 
a,  ar,  ara,  «r8,  ar4,  &c. 

The  exponent  of  r  in  the  second  term  is  1,  in  the  third  term 


PROGRESSIONS.  195 

is  2,  in  the  fourth  term  3,  and  so  on  ;  hence  the  rath  term  of  the 
series  will  be 

arn~\ 

If,  therefore,  we  put  I  for  the  last  term,  and  n  the  number  of 
terms  of  the  series,  we  shall  have 


That  is, 

The  last  term  of  a  geometrical  progression  is  equal  to  the 
product  of  the  first  term  by  that  power  of  the  ratio  whose  expo- 
nent i$.,  Qjte  less  than  the  number  of  terms. 

(242.)   To  find  the  sum  of  all  the  terms  of  a  geometrical  pro- 
gression. 

If  we  take  any  geometrical  series,  and  multiply  each  of  its 
terms  by  the  ratio,  a  new  series  will  be  formed,  of  which  ev- 
ery term  except  the  last  will  have  its  corresponding  term  in 
the  first  series.     Thus,  take  the  series 

1,  2,  4,  8,  16,  32, 
the  sum  of  which  we  will  represent  by  S,  so  that 

8=1+2+4+8  +  16+32. 
Multiplying  each  term  by  2,  we  obtain 

28=2+4+8+16+32+64. 

The  terms  of  the  two  series  are  identical,  except  the  first 
term  of  the  first  series  and  the  last  term  of  the  second  series. 
If,  then,  we  subtract  one  of  these  equations  from  the  other,  all 
the  remaining  terms  will  disappear,  and  we  shall  have 

28-8=64-1. 

In  order  to  generalize  this  method,  let  a,  ar,  ar2,  &c.,  rep- 
resent any  geometrical  series,  and  8  its  sum  ;  then 
S=a+ar+ar*+ara+  ......  +arn 

Multiplying  this  equation  by  r,  we  have 


Subtracting  the  first  equation  from  the  second,  we  obtain 
rS-S=ar"-«. 

Hence  8= — — -y-  ; 

or,  substituting  the  value  of  I  already  found,  we  shall  have 


196 


PROGRESSIONS. 


Hence,  to  find  the  sum  of  the  terms  of  a  geometrical  pro- 
gression, 

Multiply  the  last  term  by  the  ratio,  subtract  the  first  term,  and 
divide  the  remainder  by  the  ratio  less  one. 

If  the  series  is  a  decreasing  one,  and  r  consequently  repre- 
sents a  fraction,  it  is  convenient  to  change  the  signs  of  both 
numerator  and  denominator  in  this  expression,  which  then  be- 
comes 

a— ar      a—lr 

=7=r=T^r- 

(243.)  In  the  two  fundamental  equations 


s= 


Ir-a 
r-1' 


\+r.  3  -  it-i  1  J. 


•  *  'i  t* 


there  are  five  variable  quantities, 

a,  /,  r,  7i,  S, 

of  which  any  three  being  given,  the  other  two  may  be  found. 
Accordingly,  as  in  arithmetical  progression,  20  different  cases 
may  arise,  all  of  which  are  readily  solved,  with  the  exception 
of  those  in  which  n  is  the  quantity  sought.  The  value  of  n 
can  only  be  found  by  the  solution  of  an  exponential  equation. 
See  Art.  352.  These  different  cases  are  all  exhibited  in  the 
following  table  for  convenient  reference. 


No. 

Given.       |  Required. 

Formulae. 

1 

2 
3 
4 

a,  r,  n 
a,  r,  S 
#,  n,  S 
r,  7i,  S 

/ 

l=arn~\ 
a+(r-l)S 

r         ' 
/(S-/r>=a(S-arV 
(r-l)Sr"-1 

*         r"-l      ' 

5 
6 

7 
8 

a,  r,  n 
a,  r,  I 

a,  n,  / 
r,  n,  / 

S 

o     ar"-a 

=  r-l' 
Ir-a 

-$rp 

*-yf_->fif. 

"-•i/l-"-!/a  ' 
If  -I 
^-^-r-'- 

PROGRESSIONS. 


197 


No. 

Given. 

Required. 

Formulae. 

9 

a,  n,  I 

r=  v/!' 

10 

a,  n,  S 

r 

ar"-rS=a-S, 

11 

a,/,  S 

S-a 

s-r 

12 

n,/,  S 

(S-O^-Sr"-^-/. 

13 

r,  n,  I 

«=:—  j, 

14 

r,  n,  S 

a 

(r-l)S 

15 

r,  /,  S 

a=/r-(r-l)S, 

16 

n,  /,  S 

a(S~a)—  ^/(S-O—1. 

17 

a,  r,  / 

log.  I-  log.  a 

log.  r 

18 

a,  r,  S 

log.[a+(r—  1)S]  —  log.  a 

n  —                 ,        •              • 
log.r 

19 

0,  /,  S 

log.  I-  log.  a 

log.(S-a)-log.(S-l)] 

20 

r,  /,  S 

log.l—log.\lr—(i     1)S]  t  ^ 

log.r                  n- 

EXAMPLES. 

Ex.  1.  Required  the  sum  of  the  series 

1,  3,  9,  27,  &c., 
continued  to  12  terms. 


Ans.  265720. 


This  example  is  solved  by  Formula  5. 

Ex.  2.  Required  the  sum  of  the  series 

1,  2,  4,  8,  16,  &c., 

continued  to  14  terms. 

Ans.  16383. 

Ex.  3.  Given  the  first  term  2,  the  ratio  3,  and  the  number 
of  terms  10,  to  find  the  last  term. 

Ans.  39366. 

Ex.  4.  Given  the  first  term  l,the  last  term  512,  and  the  sum 
of  the  terms  1023,  to  find' the  ratio. 


198  PROGRESSIONS. 

Ex.  5.  Given  the  last  term  2048,  the  number  of  terms  12, 
and  the  ratio  2,  to  find  the  first  term. 

Ex.  6.  A  person  being  asked  to  dispose  of  his  horse,  said  he 
would  sell  him  on  condition  of  receiving  one  cent  for  the  first 
nail  in  his  shoes,  two  cents  for  the  second,  and  so  on,  doubling 
the  price  of  every  nail  to  32,  the  number  of  nails  in  his  four 
shoes.  What  would  the  horse  cost  at  that  rate  ? 

Ans.  842,949,672.95. 

(244.)  To  find  any  number  of  geometrical  means  between  two 
given  numbers. 

In  order  to  solve  this  problem,  it  is  necessary  to  know  the 
ratio.  If  m  represent  the  number  of  means,  m+2  will  be  the 
whole  number  of  terms.  Substituting  m+2  for  n  in  Formula  9 
Art.  243,  we  obtain 


That  is,  to  find  the  ratio,  divide  the  last  term  by  the  first  term, 
and  extract  the  root  denoted  by  the  number  of  means  plus  one. 

When  the  ratio  is  known,  the  required  means  are  obtained 
by  continued  multiplication. 

Ex.  1.  Find  three  geometrical  means  between  2  and  162. 

Ex.  2.  Find  two  geometrical  means  between  4  and  256. 

(245.)  Of  decreasing  progressions  having  an  infinite  number 
of  terms. 
The  formula 

a— ai* 

=T-^T' 

which  represents  the  sum  of  n  terms  of  a  decreasing  series, 
may  be  put  under  the  form 

a        ar" 
-1^~I=? 

In  a  decreasing  progression,  since  r  is  a  proper  fraction,  r" 
is  less  than  unity,  and  the  larger  the  number  n,  the  smaller  will 
be  the  quantity  r".  If,  therefore,  we  take  a  very  large  num- 
ber of  terms  of  the  series,  the  quantity  rn,  and,  consequently, 

the  term    — ,  will  be  very  small ;  and  if  we  take  n  greater 


PROGRESSIONS.  199 

arn 
than  any  assignable  number,  then  —  —  will  be  less  than  any 

assignable  number.     We  shall  therefore  have 


Hence  the  sum  of  an  infinite  series  decreasing  in  geometrical 
progression  is  found  by  the  following 

RULE. 

Divide  the  first  term  by  unity  diminished  by  the  ratio. 
Ex.  1.  Find  the  sum  of  the  infinite  series 


Here  a=l,r=$. 

Therefore,  S=-?-=-l—=2. 

Ex.  2.  Find  the  sum  of  the  infinite  series 


Ans.  f  . 
Ex.  3.  Find  the  sum  of  the  infinite  series 


Ex.  4.  Find  the  ratio  of  an  infinite  progression,  whose  first 
term  is  1,  and  the  sum  of  the  series  f. 

Ans.  }. 

Ex.  5.  Find  the  first  term  of  an  infinite  progression,  whose 
ratio  is  ~,  and  the  sum  f  . 

Ans.  | 
Ex.  6.  Find  the  first  term  of  an  infinite  progression,  of  which 

the  ratio  is  -,  and  the  sum  -  -. 
n  n—l 

PROBLEMS. 

(246.)  Prob.  1.  Of  four  numbers  in  geometrical  progression, 
the  sum  of  the  first  and  second  is  15,  and  the  sum  of  the  third 
and  fourth  is  60.  Required  the  numbers. 

Let  #,  xy,  xy*,  xy*,  be  the  numbers. 

Therefore,  x+xy  =15, 

and  xy*+xy*=QO. 


JJOO  PROGRESSIONS. 

Multiplying  the  first  equation  by  y*, 


Therefore,  yfl  =4,      ®W 

and  y=±2. 

Also,  x±2x=l5. 

Therefore,  *=5,or-15. 

Taking  the  first  value  of  x,  and  the  corresponding  value  of 
y,  we  obtain  the  series 

K      5>110'20'4fi°; 

which  numbers  may  be  easily  verified. 

Taking  the  second  value  of  x,  and  the  corresponding  value 
of  y,  we  obtain  the  series 

-15,  +30,  -60,  +120; 

which  numbers  also  perfectly  satisfy  the  problem  understood 
algebraically.  If,  however,  it  is  required  that  the  terms  of  the 
progression  be  positive,  the  last  value  of  x  would  be  inapplica- 
ble to  the  problem,  though  satisfying  the  algebraic  equation. 

Several  of  the  following  problems  also  have  two  solutions, 
if  we  admit  negative  values. 

Prob.  2.  There  are  three  numbers  in  geometrical  progres- 
sion whose  sum  is  210,  and  the  last  exceeds  the  first  by  90 
What  are  the  numbers  ? 

Ans.  30,  60,  and  120. 

Prob.  3.  There  are  three  numbers  in  geometrical  progres- 
sion whose  continued  product  is  64,  and  the  sum  of  their  cubes 
is  584.  Required  the  numbers. 

»  tabi889igof«}  gjtftot  i         mtf  fcnii^715-  2'  4>  and  8- 
Prob.  4.  There  are  four  numbers  in  geometrical  progres- 
sion, the  second  of  which  is  less  than  the  fourth  by  24  ;  and  the 
sum  of  the  extremes  is  to  the  sum  of  the  means  as  7  to  3.     Re- 
quired the  numbers. 

Ans.  1,  3,  9,  and  27. 

Prob.  5.  Of  four  numbers  in  geometrical  progression,  the 
difference  between  the  first  and  second  is  4,  and  the  difference 
between  the  third  and  fourth  is  36.  What  are  the  numbers  ? 

Ans.  2,  6,  18,  and  54. 

Prob.  6.  Of  four  numbers  in  geometrical  progression,  the 


PROGRESSIONS.  201 

sum  of  the  first  and  third  is  «,  the  sum  of  the  second  and  fourth 
is  b.     What  are  the  numbers  ? 

a3        a*b       ab*        b9 

/i  77  o 

-™' 


HARMONICA!  PROGRESSION. 

(247.)  .A  series  of  quantities  is  said  to  be  in  harmonical  pro- 
gression  when,  of  any  three  consecutive  terms,  the  first  is  to  the 
third  as  the  difference  of  the  first  and  second  is  to  the  difference 
of  the  second  and  third. 
Thus  the  numbers 

60,  30,20,  15,  12,  10, 
are  in  harmonical  progression  ;  for 

60  :  20  ::  60-30  :  30-20 
30  :  15  ::  30-20  :  20-15 
20  :  12  ::  20-15  :  15-12 
15:  10  ::  15-12:  12-10. 
So,  also,  the  numbers 

1»  J»  i»  i»  i»  J»  &c., 
form  an  harmonical  progression. 

(248.)  The  reciprocals  of  a  series  of  terms  in  harmonical  pro- 
gression form  an  arithmetical  progression. 
Thus,  the  reciprocals  of  60,  30,  20,  &c.,  are 

sV»  sV»  aV»  1*5 »  rV»  rV» 
which  are  respectively  equal  to 

eV»  /or*  ^o">  e*o>  /o">  ro» 

being  an  arithmetical  progression  whose  common  difference 
is  «V- 

If  six  musical  strings  of  equal  weight  and  tension  have  their 
lengths  in  the  ratio  of  the  numbers 

1»  !»  i>  i>  y»  J> 

.the  second  will  sound  the  octave  of  the  first ;  the  third  will 
sound  the  twelfth ;  the  fourth  will  sound  the  double  octave ; 
the  fifth  will  sound  the  eighteenth ;  and  the  sixth  will  sound 
the  third  octave  of  the  first.  Hence  the  origin  of  the  term 
harmonical  or  musical  proportion. 


202  PROGRESSIONS. 

Let  <z,  b,  c  be  three  quantities  in  harmonical  progression ; 
then 

a  :  c  : :  a—b  :  b—c; 

Sac 

whence  b— — — . 

a+c 

That  is,  an  harmonical  mean  between  two  quantities  is  equal 
to  twice  their  product  divided  by  their  sum. 


SECTION  XV. 


GREATEST  COMMON  DIVISOR.—  CONTIN- 
UED FRACTIONS.—  PERMUTATIONS  AND 
COMBINATIONS. 

(249.)  The  greatest  common  divisor  of  two  or  more  quan- 
tities is  the  greatest  factor  which  is  common  to  each  of  the 
quantities. 

THEOREM. 

,,,-*£&»- 

The  greatest  common  divisor  of  two  quantities  is  the  same 
with  the  greatest  common  divisor  of  the  least  quantity,  and  their 
remainder  after  division. 

To  prove  this  principle,  let  the  greatest  of  the  two  quantities 
be  represented  by  A,  and  the  least  by  B.  Divide  A  by  B  ; 
let  the  entire  part  of  the  quotient  be  represented  by  Q,  and  the 
remainder  by  R.  Then,  since  the  dividend  must  be  equal  to 
the  product  of  the  divisor  by  the  quotient  +  the  remainder,  we 
shall  have 


Now  every  number  which  will  divide  B  will  divide  QB  ; 
and  every  number  which  will  divide  R  and  QB  will  divide 
R+QB  or  A.  That  is,  every  number  which  is  a  common  di- 
visor of  B  and  R  is  a  common  divisor  of  A  and  B. 

Again,  every  number  which  will  divide  A  and  B  will  divide 
A  and  QB  ;  it  will  also  divide  A—  QB  or  R.  That  is,  every 
number  which  is  a  common  divisor  of  A  and  B  is  also  a  com- 
mon divisor  of  B  and  R.  Hence  the  greatest  common  divisor 
of  A  and  B  must  be  the  same  as  the  greatest  common  divisor 
of  Band  R. 


204  GREATEST    COMMON    DIVISOR. 

•  -.  '         *          k 

(250.)  To  find,  then,  the  greatest  common  divisor  of  two 
quantities,  we  divide  the  greater  by  the  less  ;  and  the  remain- 
der, which  is  necessarily  less  than  either  of  the  given  quanti- 
ties, is  by  the  last  Article  divisible  by  the  greatest  common  di- 
visor. 

Dividing  the  preceding  divisor  by  the  last  remainder,  a  still 
smaller  remainder  will  be  found,  which  is  divisible  by  the 
greatest  common  divisor  ;  and  by  continuing  this  process  with 
each  remainder  and  the  preceding  divisor,  quantities  smaller 
and  smaller  are  found,  which  are  all  divisible  by  the  greatest 
common  divisor,  until  at  length  the  greatest  common  divisor 
must  be  obtained.  Hence  the  following 

RULE. 

Divide  the  greater  quantity  by  the  less,  and  the  preceding  di- 
visor by  the  last  remainder,  till  nothing  remains  ;  the  last  divi- 
sor will  be  the  greatest  common  divisor. 

When  the  remainders  decrease  to  unity,  the  given  quanti- 
ties have  wo  common  divisor  greater  than  unity,  and  are  said 
to  be  incommensurable,  or  prime  to  each  other. 

EXAMPLES. 

Ex.  1.  What  is  the  greatest  common  divisor  of  372  and  246  ? 


372 


246 


246 

— 7~ 


246 
126 


aKvib 


126 
120 


126,  first  Remainder. 

"T 

120,  second  Remainder. 

~T 


hri 


120 
120 


6,  third  Remainder. 


20 


Here  we  have  continued  the  operation  of  division  until  we 
obtain  0  for  a  remainder ;  the  last  divisor  (6)  is  the  greatest 
common  divisor.  Thus,  246  and  372  being  each  divided  by 
6,  give  41  and  62,  and  these  quotients  are  prime  with  respect 
to  each  other ;  that  is,  have  no  common  divisor  greater  than 
unity. 


GREATEST    COMMON    DIVISOR.  205 

Ex.  2    What  is  the  greatest  common  divisor  of 
336  and  720  ? 

Ans.  48. 

Ex.  3.  What  is  the  greatest  common  divisor  of 
918  and  522? 

Ans.  18. 

(251.)  In  applying  this  rule  to  polynomials,  some  modifica- 
tion may  become  necessary.  It  may  happen  that  the  first  term 
of  the  dividend  is  not  divisible  by  the  first  term  of  the  divisor. 
This  may  arise  from  the  presence  of  a  factor  in  the  divisor 
which  is  not  found  in  the  dividend,  and  may  therefore  be  sup- 
pressed. For,  since  the  greatest  common  divisor  of  two  quan- 
tities is  only  the  product  of  their  common  factors,  it  can  not  be 
affected  by  a  factor  of  the  one  quantity  which  is  not  found  in 
the  other. 

We  may  therefore  suppress  in  the  first  polynomial  all  the 
factors  common  to  each  of  its  terms.  We  do  the  same  with 
the  second  polynomial,  and  if  the  suppressed  factors  have  a 
common  divisor,  we  reserve  it  as  forming  part  of  the  common 
divisor  sought. 

But  if,  after  this  reduction,  the  first  term  of  the  dividend, 
when  arranged  according  to  the  powers  of  some  letter,  is  not 
divisible  by  the  first  term  of  the  arranged  divisor,  we  may  mul- 
tiply the  dividend  by  any  monomial  factor  which  will  render  its 
first  term  divisible  by  the  first  term  of  the  divisor. 

This  will  not  affect  the  greatest  common  divisor,  because 
we  introduce  into  the  dividend  a  factor  which  belongs  only  to 
the  first  term  of  the  divisor ;  for  by  supposition,  all  the  factors 
common  to  each  of  its  terms  have  been  suppressed. 

EXAMPLES. 

- 

Ex.  1.  Required  the  greatest  common  divisor  of 

x*+x9  andV— 1. 
The  operation  will  here  stand  as  follows  : 

r  .  •«.»;  :..n  1  '    . . 


x'+x* 


x'-l 


x*+x,  first  Remainder. 
Suppressing  x,  we  have  x*+l. 


206  GREATEST    COMMON    DIVISOR, 


a;4-! 


x'+x* 


a;2-! 


-a:8-! 
— x9—  I. 

Whence  x*+l  is  the  greatest  common  divisor.  To  verify 
this  result,  divide  x*+x9  by  #a+l,  and  we  obtain  x* ;  divide 
x*—l  by  x*-\-l,  and  we  obtain  x2— 1. 

Ex.  2.  Required  the  greatest  common  divisor  of 

x*-b*x  and  x*+2bx+b\ 

Suppressing  the  factor  x  in  the  first  polynomial,  we  proceed 
as  follows : 


2bx+2b*,  first  Remainder. 
Suppressing  the  factor  2b, 

za-&2  x+b 


x*+bx 


x-b 


\*  «t>  im;^ 

7  7  « 

— bx— &3 

Whence  x-t-b  is  the  greatest  common  divisor. 
Ex.  3.  Required  the  greatest  common  divisor  of 
4a3-2aa-3a+l  and  3aa-2a-l. 

Ans.  a— I. 

Ex.  4.  Find  the  greatest  common  divisor  of 
x*—a*  and  x*—a*. 

Ans.  x—a 

Ex.  5.  Find  the  greatest  common  divisor  of 
<za-3a&+2fca  and  aa-a&-2&2. 

Ans.  a—2b. 

Ex.  6.  Find  the  greatest  common  divisor  of 
a*—x*  and  a8— cfx— ax*-t-x*. 

Ans.  a'-x* 
Ex.  7.  Find  the  greatest  common  divisor  of 

as-aafe+3a&2-3&8  and  a'-5ab+4b\ 

Ans.  a—b 
Ex.  8.  Find  the  greatest  common  divisor  of 

ci'-3ab+ac+2b*-2bc  and  a'-b*+2bc-c\ 


CONTINUED  FRACTIONS.  207 


CONTINUED  FRACTIONS. 

(252.)  From  the  operation  on  page  204,  we  see  that  the 

,      ..      246.  .1 

fraction  —  is  equal  to 


Also,  the  fraction  ——  is  equal  to 


246  1-Hff 


Therefore,  —  is  equal  to 


Again,  the  fraction  T^  is  equal  to      •        or  . 

A<*O  *   i  T¥¥  *•  +  ¥7 

Therefore,  -  -  is  equal  to 


372  1  +  1 


1  +  1 


1+3*0, 

which  is  called  a  continued  fraction. 

A  continued  fraction  is  one  whose  numerator  is  unity,  and  its 
denominator  an  integer  plus  a  fraction  whose  numerator  is  like- 
wise unity,  and  its  denominator  an  integer  plus  a  fraction,  and 
so  on. 

The  general  form  of  a  continued  fraction  is 


_ 
&+ 


_ 
e+1,  &c. 

(253.)  Any  fraction  may  be  transformed  into  a  continued 
fraction  by  the  method  of  finding  the  greatest  common  divisor 
of  the  numerator  and  denominator. 

114 

Ex.  1.  Transform  -—  into  a  continued  fraction. 

O'l  i 

An*.  1_ 

3+1  _ 
22  +  l 


10 


208  CONTINUED    FRACTIONS. 

Off  I 

Ex.  2.  Transform  r^-  into  a  continued  fraction. 
Uoo 

Ans.  l__ 


2+1 

.— TT.tl*  ! 


2+1  _ 

1  +  aV 

421 

Ex.  3.  Transform  —  into  a  continued  fraction. 

251 
Ex.  4.  Transform  —  -  into  a  continued  fraction. 

130 

Ex.  5.  Transform  —  into  a  continued  fraction. 

(254.)  The  value  of  a  continued  fraction,  when  composed  ot 
a  finite  number  of  terms,  is  easily  found. 

Ex.  1.  Find  the  value  of  the  continued  fraction 

ft   ^tOJ 

2+T 


Beginning  with  the  last  fraction,  we  have 

" 


„  1  4 

Hence  ^r-r-T=— . 

lo 


1  30 

Therefore,  2+I+l=l3- 

And  i—          13 

2+^T=^  ^ 

o-t-T 

fe.  2.  Find  the  value  of  the  continued  fraction 
1 


3  +  1 


.  3.  Find  the  value  of  the  continued  fraction 


CONTINUED    FRACTIONS.  209 


3+1 


(255.)  When  a  fraction  has  been  transformed  into  a  con- 
tinued fraction,  its  approximate  value  may  be  found  by  taking 
a  few  of  the  first  terms  of  the  continued  fraction. 

114 

Thus,  an  approximate  value  of  —  -  is  i,  which  is  the  first 

term  of  its  continued  fraction. 

22 

By  taking  two  terms,  we  obtain  —  ,  which  is  a  nearer  ap- 

proximation ;  and  three  terms  would  give  a  still  more  accurate 
~alue. 

532 

Ex.  1.  Find  approximate  values  of  the  fraction 


1   4  33 
-,-.-. 


Ex.  2.  Find  approximate  values  of  the  fraction  ~-. 


119 
Ex.  3.  Find  approximate  values  of  the  fraction  -—  :. 

(256.)  By  this  method  we  are  enabled  to  discover  the  ap- 
proximate value  of  a  fraction  expressed  in  large  numbers  ;  and 
this  principle  has  some  important  applications,  particularly  in 
Astronomy. 

Ex.  4.  The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  3.1415926.  Find  approximate  values  for  this 
ratio. 

22    333    355 
7'  106'  113* 

Ex.  5.  In  87969  years,  the  Earth  makes  277287  conjunc- 
tions with  Mercury.     Find  approximate  values  for  the  frac- 
87969 
27728T 

1    6     7    13    33 
AnS'  3'  19'  22'  4?  104' 


210  PERMUTATIONS    AND    COMBINATIONS. 

Ex.  6.  In  57551  years,  the  Earth  makes  36000  conjunctions 
with  Venus.  Find  approximate  values  for  the  fraction  ^  . 

8  235 
An,.-,—. 

Ex.  7.  In  295306  years,  the  Moon  makes  3652422  synod- 
ical  revolutions.  Find  an  approximate  value  of  the  fraction 
295306 

3652422* 

19 

AnS-  235' 

THEORY  OF  PERMUTATIONS  AND  COMBINATIONS. 

(257.)  The  different  orders  in  which  quantities  may  be  ar- 
ranged are  called  their  Permutations.  Thus, 


the  permutations  of  the  three  letters  a,  b,  c,  taken  all, 
together,  are 

•  ;•;,..  -*l« 


•-  r.fo?K*>j3*fi  -:f?fi7  ^AfEfhroittcri)  bn5/i3 

The  permutations  of  the  same  letters  taken  two  and 
two,  are    ......      %•  •.,     • ..- 

•  ulu1..   >>).;t--     ii;    i-'SBp^^lDKQ  I''.'!  .U^IJ'il  2>  *(.*  oiil.s^  J 

'!'    '' 


The  permutations  of  the  same  letters  taken  singly,  or  \  , ' 

one  by  one,  are       f.?'f  IX^  ^.  ^     .         .         .  )    » 

'  c. 

(258.)  To  find  the  number  of  permutations  of  n  toers,  taken 
m  flTid  m  together. 

Let  a,b,c,d A,  be  the  n  letters. 

The  number  of  permutations  of  n  letters  taken  singly,  or  one 
by  one,  is  evidently  equal  to  the  number  of  letters,  or  to  n. 

The  number  of  permutations  of  n  letters  taken  two  and  two 
is  n(n—  1).  For  if  we  reserve  one  of  the  letters,  as  a,  there 
will  remain  n—l  letters, 


PERMUTATIONS    AND   COMBINATIONS.  211 

b,  C,  d k. 

Writing  a  before  each  of  these  letters,  we  shall  have 

ab,  ac,  ad dk  ; 

that  is,  we  obtain  n—  1  permutations  of  the  n  letters  taken  two 
and  two,  in  which  a  stands  first.  Proceeding  in  the  same 
manner  with  6,  we  shall  find  n—  1  permutations  of  the  n  letters 
taken  two  and  two,  in  which  b  stands  first ;  and  so  for  each 
of  the  n  letters.  Hence  the  whole  number  of  permutations 
will  be 

n(n-l). 

The  number  of  permutations  of  n  letters  taken  three  and 
three  together  is 

n(n-l)  (n-2). 

For  if  we  reserve  one  of  the  letters,  as  a,  there  will  remain 
,i—l  letters.     Now  we  have  found  the  number  of  permuta- 
tions of  n  letters  taken  two  and  two  to  be  71(71—!).     Hence 
the  permutations  of  n—  1  letters  taken  two  and  two  must  be 
(n-l)  (n-2). 

Writing  a  before  each  of  these  permutations,  we  shall  have 
(n—  1)  (n— 2)  permutations  of  the  n  letters  taken  three  and 
three,  in  which  a  stands  first.  Proceeding  in  the  same  manner 
with  bj  we  shall  find  (n—l)  (n—2)  permutations  of  the  n  let- 
ters taken  three  and  three,  in  which  b  stands  first ;  and  so  for 
each  of  the  n  letters.  Hence  the  whole  number  of  permuta- 
tions will  be 

n(n-l)  (n-2). 

In  like  manner,  we  can  prove  that  the  number  of  permuta- 
tions of  n  letters  taken  four  and  four  is 

n(n-l)  (n-2)  (n-3). 

When  the  letters  are  taken  two  and  two,  the  last  factor  in 
the  formula  representing  the  number  of  permutations  is  n—  1. 
When  the  letters  are  taken  three  and  three,  the  last  factor  is 
n—2.  When  the  letters  are  taken  four  and  four,  the  last 
factor  is  n—3. 

Hence,  when  the  letters  are  taken  m  and  m  together,  the  last 
factor  will  be  n—  (m—  1)  or  n— m+1;  and  the  number  of  per- 
mutations of  n  letters  taken  m  and  m  together  will  according- 
ly be 

n(n-l)  (n-2)  (n-3) (n-m+l). 


212  PERMUTATIONS    AND    COMBINATIONS. 


EXAMPLES. 

Ex.  1.  Required  the  number  of  permutations  of  the  8  letters 
a,  6,  c,  d,  e,f,  g,  h,  taken  5  and  5  together. 

Here  w=8,  m=5,  n—m+l=4, 

and  the  above  formula  becomes 

8.7.6.5.4=6720,  Ans. 

Ex.  2.  Required  the  number  of  permutations  of  the  26  let- 
ters of  the  alphabet,  taken  4  and  4  together. 

Ans.  358800. 

Ex.  3.  Required  the  number  of  permutations  of  12  letters, 
taken  6  and  6  together. 

Ans.  665280. 

(259.)  If  we  suppose  that  each  permutation  comprehends  all 
the  n  letters  ;  that  is,  if  m=n,  the  preceding  formula  becomes 

w(rc-l)  (n-2) 2X1; 

or,  inverting  the  order  of  the  factors, 

1.2.3.4 (n-l)n; 

which  expresses  the  number  of  permutations  of  n  letters  taken 
all  together. 

Ex.  1.  Required  the  number  of  changes  which  can  be  rung 
upon  8  bells. 

According  to  the  preceding  formula,  we  have 

1.2.3.4.5.6.7.8=40320,  Ans. 

Ex.  2.  How  many  permutations  may  be  formed  from  the 
letters  of  the  word  Roma  ? 

Ex.  3.  What  is  the  number  of  permutations  which  may  be 
formed  from  the  letters  composing  the  word  "  virtue  ?" 

Ex.  4.  What  is  the  number  of  different  arrangements  which 
can  be  made  of  12  persons  at  a  dinner-table  ? 

Ans.  479001600. 

(260.)  The  combinations  of  any  number  of  quantities  signify 
the  different  collections  which  may  be  formed  of  these  quanti- 
ties, without  regard  to  the  order  of  their  arrangement. 

Thus,  the  three  letters  a,  6,  c,  taken  all  together,  form  but 
one  combination,  abc. 

Taken  two  and  two,  they  form  three  combinations, 
ab,  ac,  be. 


PERMUTATIONS    AND    COMBINATIONS.  213 

(261.)  To  find  the  number  of  combinations  of  n  letters,  taken 
m  and  m  together. 

The  number  of  combinations  of  n  letters  taken  separately,  or 
one  by  one,  is  evidently  n. 

The  number  of  combinations  of  n  letters  taken  two  and  two, 

«(n-l) 
1.2 

For  the  number  of  permutations  of  n  letters  taken  two  and 
two  is  n(n—  1)  ;  and  there  are  two  permutations  (ab,  ba)  cor- 
responding to  one  combination  of  two  letters.  Therefore  the 
number  of  combinations  will  be  found  by  dividing  the  number 
of  permutations  by  2. 

The  number  of  combinations  of  n  letters  taken  three  and 

n(n-l)  (7i-2) 
three  together,  is  —  -  -  -—-5  -  -. 


For  the  number  of  permutations  of  n  letters  taken  three  and 
three,  is  n(n—  1)  (n—  2)  ;  and  there  are  1.2.3  permutations  for 
one  combination  of  three  letters.  Therefore  the  number  of 
combinations  will  be  found  by  dividing  the  number  of  permu- 
tations by  1.2.3. 

In  the  same  manner,  we  shall  find  the  number  of  combina- 
tions of  n  letters,  taken  m  and  m  together,  to  be 
n(n-l)  (n-2)  .....  (n-m+l) 

1.2.3          .....          m 

Ex  i.  Required  the  number  of  combinations  of  six  letters 
taken  three  and  three  together. 

Here  71=6,  m—  3,  n—  wz+l=4, 

and  the  formula  becomes 

6.5.4 

u^T20- 

Ex.  2.  Required  the  number  of  combinations  of  8  letters 
taken  4  and  4. 

Ans.  70. 

Ex.  3.  Required  the  number  of  combinations  of  10  letters 
taken  6  and  6. 

Ans.  210. 

The  following  table,  which  is  computed  by  the  preceding  for- 
mula, shows  the  number  of  combinations  of  1,  2,  3,  4,  &c.,  let- 


214 


PERMUTATIONS    AND   COMBINATIONS. 


ters  taken  singly,  or  two  and  two,  three  and  three,  &c.  An 
important  application  of  these  principles  will  be  seen  in  the  next 
Section. 


Letters. 

Singly. 

2  and  2. 

3  and  3. 

4  and  4. 

5  and  5. 

6  and  6. 

7  and  7. 

8  and  8. 

9  and  9.  [10  and  10. 

1 

1 

2 
3 

2 
3 

1 

3 

1 

Number  of  combinations. 

4 

4 

6 

4 

1 

5 

5 

10 

10 

5 

1 

6 

6 

15 

20 

15 

6 

1 

7 

7 

21 

35 

35 

21 

7 

1 

8 

8 

28 

56 

70 

56 

28 

8 

1 

9 

9 

36 

84 

126 

126 

84 

36 

9 

1 

10 

10 

45 

120 

210 

252 

210 

120 

45 

10 

1 

ia.  ':; 


SECTION  XVI. 


INVOLUTION  OF  BINOMIALS. 

(262.)  We  have  shown,  in  Art.  142,  how  to  obtain  any 
power  of  a  binomial  by  actual  multiplication.  We  now  pro- 
pose to  develop  a  theorem  by  which  this  labor  may  be  greatly 
abridged. 

Taking  the  binomial  a+b,  its  successive  powers  found  by 
actual  multiplication  are  as  follows  : 
(a+b)l=a  +6, 
(a+by=ai+2ab  +b\ 


The  powers  of  a—  b,  found  in  the  same  manner,  are  as  fol- 
lows : 

(a-b)l=a  -b, 
(a-by=a*-2ab  -f&2, 


On  comparing  the  powers  of  a+b  with  those  of  a—  b,  we 
perceive  that  they  only  differ  in  the  signs  of  certain  terms.  In 
the  powers  of  a+b,  all  the  terms  are  positive.  In  the  powers 
of  a—  5,  the  terms  containing  the  odd  powers  of  b  have  the 
sign—,  while  the  even  powers  retain  the  sign  +.  The  reason 
of  this  is  obvious  ;  for,  since  —b  is  the  only  negative  term  of 
the  root,  the  terms  of  the  power  can  only  be  rendered  nega- 

10* 


216  INVOLUTION    OF    BINOMIALS. 

tive  by  b.  A  term  which  contains  the  factor  —b  an  even 
number  of  times,  will  therefore  be  positive ;  if  it  contain  it  an 
odd  number  of  times,  it  must  be  negative.  Hence  it  appears 
that  it  is  only  necessary  to  seek  for  a  method  of  obtaining  the 
powers  of  a+b ;  for  these  will  become  the  powers  of  a— b  by 
simply  changing  the  signs  of  the  alternate  terms. 

(263.)  If  we  consider  the  exponents  of  the  preceding  pow- 
ers, we  shall  find  that  they  follow  a  very  simple  law.     Thus, 


of  a  are   2,  1,0, 
of  b  are   0,  1,2. 


In  the  square,  the  exponents  .     .    j 

(  of  a  are   3,2,1,0, 
In  the  cube,  the  exponents       .     . 

(  of  b  are   0,  1,  2,  3. 

(  of  a  are   4,  3,2,  1,  0, 
In  the  fourth  power,  the  exponents  ]    c  .          .   ,   _  0 

(  of  b  are   0,  1,  2,  3,  4. 

&c.,  &c.,  &c. 

In  the  first  term  of  each  power,  a  is  raised  to  the  required 
power  of  the  binomial ;  and  in  the  following  terms,  the  expo- 
nents of  a  continually  decrease  by  unity  to  0 ;  while  the  ex- 
ponents of  b  increase  by  unity  from  0  up  to  the  required  power 
of  the  binomial.  It  is  obvious  that  this  will  always  be  the  case, 
to  whatever  extent  the  involution  may  be  carried.  Also,  the 
sum  of  the  exponents  of  a  and  b  in  any  term  is  equal  to  the  ex- 
ponent of  the  power  required.  Thus,  in  the  second  power,  the 
sum  of  the  exponents  of  a  and  b  in  each  term  is  2  ;  in  the  third 
power  it  is  3 ;  in  the  fourth  power,  4,  &c. 

We  hence  infer,  that  for  the  seventh  power  the  terms,  with- 
out the  coefficients,  must  be 

a\  a*b,  a*V,  ctb\  «"&*,  a*b\  ab\  V; 
and  for  the  nth  power, 

a",  an~lb,  an~*b\  an~3b* a'b"-*,  abn~l,  b\ 

(264.)  It  remains  to  determine  the  coefficients  which  belong 
to  these  terms  ;  and  in  order  to  discover  the  law  of  their  forma- 
tion, let  us  take  the  coefficients  already  found  by  themselves. 
The  coefficients  of  the  1st  power  are  1       1 

"  2d        "  121 

"  3d        "  1331 

"  4th       "  14641 

"  5th       «  1    5     10     10     5    1 

"  6th       "  1    6    15    20    15    6    1 


INVOLUTION    OF    BINOMIALS.  217 

The  numbers  in  this  table  are  identical  with  those  in  the  ta- 
ble of  combinations  on  page  214.  For  example,  the  coefficients 
of  the  fifth  power  denote  the  number  of  combinations  of  five 
letters  taken  one  and  one,  two  and  two,  &c. ;  the  coefficients 
of  the  sixth  power  denote  the  number  of  combinations  of  six 
letters  taken  one  and  one,  two  and  two,  &c.  The  reason  of 
this  will  appear  if  we  observe  the  law  of  the  product  of  several 
binomial  factors,  x+a,  x+b,  x+c,  x+d,  &c. 

Multiplying        x  +  a 
by  x  +  b, 

we  obtain  x*+(a+b)x+ab=lst  product. 

Multiplying  by  x  +  c, 

we  obtain  x*+ (a+b  +  c)x*  +  (ab  +  ac  +  bc)x  +  abc  =  2d 

product. 

Multiplying  by  x  +  d, 

we  obtain  x*  +  (a+b+c+d)x*  +  (ab+ac+ad+bc  +  bd+ 

cd)x*  +  (abc + abd+  acd+ bcd)x + abcd=  3d  product. 

We  observe  that  in  each  of  these  products  the  coefficient  of 
x  in  the  first  term  is  unity ;  the  coefficient  of  the  second  term  is 
the  sum  of  the  second  terms  of  the  binomial  factors  ;  the  coefficient 
of  the  third  term  is  the  sum  of  all  their  products  taken  two  and 
two ;  the  coefficient  of  the  fourth  term  is  the  sum  of  all  their 
products  taken  three  and  three,  &c. 

It  is  easily  seen  that  if  we  multiply  the  last  product  by  a 
new  factor,  x+e,  the  same  law  of  the  coefficients  will  be  pre- 
served. Hence  the  law  is  general. 

If  now,  in  the  preceding  binomial  factors,  we  suppose  «,  b, 
c.  d,  &c.,  to  be  all  equal  to  each  other,  the  product 

(x+a)  (x+b)  (x+c)  (x+d) 

becomes  (x+a)n. 

The  coefficient  of  the  second  term  of  the  product,  or  a+b-\- 

c+d ,  becomes  a+a+a-\-a ;  that  is,  a  taken  as 

many  times  as  there  are  letters  a,  b,  c,  d,  and  is,  consequently, 
equal  to  na. 

The  coefficient  of  the  third  term,  or  ab+ac,  &c.,  reduces  to 

aa_l_aa_l_a9 9  or  a2  repeated  as  many  times  as  there  are 

different  combinations  of  n  letters  taken  two  and  two ;  that  is, 

n(n—l)  , 
hv  Art.  261,  to     v         '  a\ 


218  INVOLUTION    OF    BINOMIALS. 

The  coefficient  of  the  fourth  term  reduces  to  a3  repeated  as 
many  times  as  there  are  different  combinations  of  n  letters 

n(n-l)  (n—2)  , 
taken  three  and  three  ;  that  is,  —  -  —  j-^-~  -  a3,  and  so  on. 

Thus  we  find  that  the  nth  power  of  x+a  may  be  expressed 
as  follows  : 

•  •  • 


which  is  called  the  BINOMIAL  FORMULA,  and  is  generally  as- 
cribed to  Sir  Isaac  Newton.  So  important  was  it  regarded, 
that  it  was  engraved  on  his  monument  in  Westminster  Abbey 
as  one  of  his  greatest  discoveries. 

On  comparing  the  different  terms  of  this  development,  we 
perceive  that  any  coefficient  may  be  derived  from  the  preced- 
ing one  by  the  following  rule  :  If  the  coefficient  of  any  term  be 
multiplied  by  the  exponent  of  x  in  that  term,  and  divided  by  the 
exponent  of  a  increased  by  one,  it  will  give  the  coefficient  of  the 
succeeding  term. 

Thus,  the  fifth  power  of  x+a  is 


If  the  coefficient  5  of  the  second  term  be  multiplied  by  4, 
the  exponent  of  x  in  that  term,  and  divided  by  2,  which  is  the 
exponent  of  a  increased  by  one,  we  obtain  10,  the  coefficient 
of  the  third  term. 

So,  also,  if  10,  the  coefficient  of  the  fourth  term,  be  multi- 
plied by  2,  the  exponent  of  x,  and  divided  by  4,  the  exponent 
of  a  increased  by  one,  we  obtain  5,  the  coefficient  of  the  fifth 
term  ;  and  so  of  the  others. 

The  coefficients  of  the  sixth  power  will  also  be  found  as  fol- 
lows: 

6X5   15X4   20X3    15X2   6X1 
'    '     2    '       3~~'      4     '~5~~'  ~6~; 
that  is,          1,6,    15,        20,        15,         6,         1. 

The  coefficients  of  the  seventh  power  will  be 

7  7x6  21x5  35X4  35X3  21X2  7X1 

2    '      3     '      4     '  ~~5~'  ~~6~'  ~T~  5 
that  is,     1,7,    21,      35,        35,        21,         7,         1. 


INVOLUTION    OF    BINOMIALS.  219 

Therefore,  the  seventh  power  of  x  +a  is 


It  is  sometimes  preferable  to  retain  the  factors  of  the  coeffi- 
cients distinct  from  each  other,  as  follows  : 


7.6.5.4.3  §  a     7.6.5.4.3.2  6       7.6.5.4.3.2.1   7 
1.2.3.4.5^  X  +1.2.3.4.5.6ff  ;  +  1.2.3.4.5.6/7*  ' 

The  factor  1  is  retained  for  the  sake  of  symmetry,  and  to 
exhibit  more  clearly  the  law  of  the  coefficients. 
(265.)  The  following,  therefore,  is  the 

BINOMIAL  THEOREM. 

In  any  power  of  a  binomial  x+a,  the  exponent  of  x  begins  in 
the  first  term  with  the  exponent  of  the  power,  and  in  the  follow- 
ing terms  continually  decreases  by  one.  The  exponent  of  a  com- 
mences with  one  in  the  second  term  of  the  power,  and  continually 
increases  by  one. 

The  coefficient  of  the  first  term  is  one  ;  that  of  the  second  is 
the  exponent  of  the  power  ;  and  if  the  coefficient  of  any  term  be 
multiplied  by  the  exponent  of  x  in  that  term,  and  divided  by  the 
exponent  of  a  increased  by  one,  it  will  give  the  coefficient  of  the 
succeeding  term. 

(266.)  The  number  of  terms  in  the  power  is  always  greater 
by  unity  than  the  exponent  of  the  power.  Thus,  the  number 
of  terms  in  (a+b)*  is  4+1,  or  5  ;  in  (a+b)6  is  6  +  1,  or  7. 

Also,  if  we  examine  the  table  in  Art.  264,  it  will  be  per- 
ceived that,  after  we  pass  the  middle  term,  the  same  coeffi- 
cients are  repeated  in  the  inverse  order.  Thus,  the  coeffi- 
cients of 

(a+b)*  are  1,  5,  10,  10,  5,  1  ; 

of     (a+b)°  are  1,  6,  15,  20,  15,  6,  1. 

Hence  it  is  only  necessary  to  compute  the  ^coefficients  for 
half  the  terms  ;  we  then  repeat  the  same  numbers  in  the  in- 
verse order. 

(267.)  The  sum  of  the  coefficients  for  each  power  is  equal  to  the 
number  2  raised  to  the  same  power.  For,  let  x=l  and  a=l, 
then  each  term  without  the  coefficients  reduces  to  unity,  and 


220  INVOLUTION    OF    BINOMIALS. 

the  value  of  the  power  is  simply  the  sum  of  the  coefficients. 
.Also,  in  this  case,  (x+a)n  becomes  (l  +  l)w,  or  2n.     Thus  the 
coefficients  of  the 
first  power  are     1  +  1  =2=2*  ; 
second      "  l+2+l=4=2a;  v% 

third          "  1+3+3+1  =8=2"; 

fourth        "  1  +4+6+4+  1  =  16=24, 

&c.,  &c.,  &c. 

EXAMPLES. 

Ex.  1.  Raise  x+a  to  the  9th  power. 
The  terms  without  the  coefficients  are 

xg,  ax6,  a?x\  a*xe,  a*x*9  cfx*,  a'x*,  cfx*,  a*x,  a9. 
And  the  coefficients  are 

9X8  36X7  84X6   126X5   126X4  84X3  36X2  9X1 
lf  °'  ~1T'  "3"'  "IT"'  ~~5~'  ~~6~~'  ~T~'  ~~8~'  ~9~' 
that  is, 

1,  9,    36,       84,        126,       126,         84,         36,          9,          1, 
Prefixing  the  coefficients,  we  obtain 


It  should  be  remembered  that,  according  to  Art.  266,  it  is 
only  necessary  to  compute  the  coefficients  of  half  the  terms  in- 
dependently. 

Ex.  2.  What  is  the  6th  power  ofx—a? 

(268.)  If  the  terms  of  the  given  binomial  are  affected  with 
coefficients  or  exponents,  they  must  be  raised  to  the  required 
powers,  according  to  the  principles  already  established  for  the 
involution  of  monomials. 

Ex.  3.  Raise  2#+5a3  to  the  fourth  power. 

For  convenience,  let  us  substitute  b  for  2x,  and  c  for  5a\ 

Then  (6+c)4=64+468c+66V+4&cs+c4. 

Restoring  the  values  of  b  and  c, 

The  first  term  will  be     (2x)*  =I6x*. 

The  second  term    "     4(2x)3X  5<z3  =4.8.5#V. 

The  third  term       «      6(249X(5aa)a=6.4.25a:V. 

The  fourth  term     "      4(2z)  X(5a2)3=4.2.125z<z'. 

The  fifth  term         "  (5«9)4=625«8. 


INVOLUTION    OF    BINOMIALS.  221 

Therefore, 


Ex.  4.  What  is  the  fourth  power  of  2x*+4y*? 
Ex.  5.  What  is  the  seventh  power  of  2a—  3b  ? 

Arts.  128a7-134406&+6048a6&2-15120a4&3+22680a3&4 
-20412a2&5+10206a&8-2187&7. 
Ex.  6.  What  is  the  sixth  power  of  a*+3ab  ? 

Ans.  a16  +  1  8al*b  +  1  3501463  +  540«1263  +121  5a1064  + 


Ex.  7.  What  is  the  fifth  power  of  5c2-4y2z  ? 

(269.)  By  means  of  the  Binomial  Theorem  we  can  raise  any 
polynomial  to  any  power. 

For  example,  let  it  be  required  to  raise  a+b+c  to  the  third 
power. 

For  convenience,  we  put  b+c=m;  we  then  have 


Substituting  for  m  its  equal,  &+c,  we  obtain 


We  must  now  develop  the  powers  of  the  binomial  b+c,  and 
perform  the  multiplications  which  are  here  indicated.  We 
thus  obtain 


+c3. 

Ex.  2.  Raise  x+a+b  to  the  fifth  power. 
(270.)  When  one  of  the  terms  of  a  binomial  is  unity,  the 
powers  assume  a  simpler  form,  since  every  power  of  1  is  1. 
Thus,  the  fourth  power  of  a+ft,  which  is 


when  we  make  0=1,  becomes 


n      n(n—l)  ,    n(n—  1)  (n—  2)  , 
So,  also,  (l+g)-=l+-g+   \2  V+  --  £j±  -  V+,&c. 

Every  binomial  of  the  form  (x+a)n  may  be  reduced  to  the 

(a\n 
1+-  J  .     For 


INVOLUTION    OF    BINOMIALS. 


x+a   =; 


Therefore,  (x+a)n=xn  (l  +-Y, 


This  expression  for  the  value  of  (x+a)n  is  equivalent  to  that 
on  page  218,  as  may  be  easily  shown  by  multiplying  xn  into 
each  term  within  the  parenthesis.  For  some  purposes  this  is 
regarded  as  the  simplest  form. 

(271.)  In  the  development  of  the  binomial  (x+a)n,  we  have 
hitherto  supposed  n  to  be  a  positive  integer.  The  Binomial 
Theorem  is,  however,  applicable,  whatever  be  the  nature  of 
the  quantity  n,  whether  it  be  positive  or  negative,  integral  or 
fractional.  When  n  is  a  positive  integer,  the  series  consists 
of  Ti+1  terms.  In  every  other  case,  the  series  never  termin- 
ates ;  that  is,  the  development  of  (x+a)n  furnishes  an  infinite 
series. 

Ex.  1.  It  is  required  to  convert  —  r  or  (a+b)~l  into  an  in- 

finite series. 

According  to  Art.  265,  the  terms  without  the  coefficients  are 
a~\  a-*b,  a~5b*,  a~*b\  aT*b\  a~«b\  ar7be,  &c. 

The  coefficient  of  the  first  term  is  1. 

The  coefficient  of  the  second  term  is  —  1,  the  exponent  of 
the  power. 

_  1  vy  _  O 

The  coefficient  of  the  third  term  is      -  -  --  =  +  1. 
fourth    « 

fifth        « 
sixth      . 


We  thus  obtain 

,  &c., 


where  the'  law  of  the  series  is  obvious  ;  the  coefficients  are  all 
unity,  and  the  signs  are  alternately  positive  and  negative. 


INVOLUTION    OP    BINOMIALS. 


223 


We  might  have  obtained  the  same  result  by  the  ordinary 
method  of  division.     The  operation  is  as  follows  : 


a+b 


5! 

"a8' 


£ 
a4" 


=  the  quotient. 


Hence, 

1       1     b     &'     b*   t 

—  —  r=  ---  H  —  5  —  i,  &c.,  which  may  be  written 
a+b    a    a*    a*     a* 

a~l—cr*b+a~3b*—a~4b*+,&,c.,  the  same  as  before  found; 

and  it  is  obvious,  from  inspecting  the  operation  of  division,  that 
the  series  will  never  terminate. 

Ex.  2.  It  is  required  to  convert         ,.a  or  (a+b)~*  into  an 


f      ,. 

1     26  ,  3&9    4&8    5&4 
AnS'      -+"     +-~' 


infinite  series. 


or,  «-2-2a-36+3a-4&3-4a-5Z>3+5a-664,  &c. 

Here  the  coefficients  increase  regularly  by  1,  and  the  signs 
are  alternately  positive  and  negative.  We  might  have  ob- 
tained the  same  result  by  division,  as  in  the  former  example. 

Ex.  3.  Expand  into  a  series  •—  r  or  (a—  6)"1. 

Here  the  coefficients  furnished  by  the  Rule  are 
+  1,  -1,  +1,  -1,  &c. 

But  the  factor  b  being  negative,  all  its  odd  powers  are  nega- 
tive. Hence  the  second  term  contains  two  negative  factors, 
so  that  its  resulting  sign  is  +.  The  same  remark  applies  to 


224  INVOLUTION    OF    BINOMIALS. 

the  fourth  and  sixth  terms,  &c.,  making  the  terms  of  the  series 
all  positive. 

Ex.  4.  Expand  into  a  series  ,       ,.a  or  (a—b)~*. 
Ex.  5.  Expand  into  a  series  (a+b)~*. 

Ex.  6.  Expand  into  a  series  (a— b)~*. 

(272.)  We  have  now  considered  the  powers  of  a  binomial 
when  the  exponent  is  an  integer,  either  positive  or  negative. 
It  remains  to  consider  the  case  when  the  exponent  is  a.  fraction. 

EXAMPLES. 

Ex.  I.  Expand  Va+b  or  (a+b)*  into  an  infinite  series. 
The  terms  without  the  coefficients  are 

The  exponents  of  a  decrease  by  unity,  while  those  of  b  in- 
crease by  unity. 

The  coefficient  of  the  first  term  is  1. 

"  second  "  -J — . 

2          ~2A' 
fourth    -          -^=+S 

"  fifth 


4  2.4.6.8* 

The  series,  therefore,  is 


The  factors  which  form  the  coefficients  are  kept  distinct,  in 
order  to  show  more  clearly  the  law  of  the  series.  The  numer- 
ators of  the  coefficients  contain  the  series  of  odd  numbers,  1,3, 
5,  7,  &c.,  while  the  denominators  contain  the  even  numbers, 
2,  4,  6,  8,  10,  &c, 

The  above  series  expresses  the  square  root  of  a+b.  We 
shall  obtain  the  same  result  if  we  extract  the  square  root  by 
the  usual  method.  See  Art.  299. 


INVOLUTION    OF    BINOMIALS.  225 

Ex.  2.   It  is  required  to  convert  (a*+x)2  into  an  infinite 
series. 

3.5.QTV     3.5.7a~V 
2.4.6.8  +  2.4.6.8.10        ' 
x_          x*          3x*  3.5x*  3.5.7x5 

+2^    ~25^+2Z6^~2Z6^+2.4.6.8.10a9     '& 
where  the  law  of  the  series  is  evident. 

i 
Ex.  3.  It  is  required  to  convert  (a—x)2  into  an  infinite 

series. 

Ex.  4.   It  is  required  to  convert  (a+b)^  into  an  infinite 
series. 

b       263        2.563         2.5.S64 


Ex.  5.  Expand  (a— b)*  into  an  infinite  series. 

T  Ql.2  Q    fy7i3  Q    T    1  1  A* 

*  \   ,        0          00             d.70              6. 7.110  . 

Ji.715.    a     1   1 2 o 1 — ;  &C.  t   . 

4<z     4.8<z      4.8.12<z      4.8. 12.16fl 
.Zftr.  6.  Expand  (a+x)^  into  an  infinite  series. 

Ex.  7.  Expand  (1— x)7  into  an  infinite  series. 

1         1.4    2       1.4.9     3       1.4.9.14     4_ 
Ans.  1  -g^-^Q^  ~ 5. 10. 15^  "5.10.15.20^      ' 

Ex.  8.  Expand  (a3— 03)¥  into  an  infinite  series. 

/        b3       2b*        2.569 
JL715.  a[  1—  — -^-— — -Q—  ——-f -g— ,  <fe 


(273.)  The  binomial  theorem  is  also  applicable  to  cases  in 
which  the  value  of  the  exponent  n  is  a  negative  fraction. 


EXAMPLES. 


JBa;.  1.  Expand  into  a  series  i  or  (a+b)~%. 

(a+b)2 

The  terms  without  the  coefficients  are 


The  coefficient  of  the  first  term  is         1. 
"  second  "          —  -. 

P 


226  INVOLUTION    OF   BINOMIALS. 

—  —  ^  —  -2.  1   Q 

The  coefficient  of  the  third  term  is        — -= H — — . 

<w  /w«  * 

HV         5  1    Q  K 

«  /.  .1  .,  24-^  a  1.0. 0 

fourth    «  -^^=___. 


fifth 
Hence  we  obtain 


&C. 


2.  Expand  into  an  infinite  series  (a+x)~~3. 

_i     1  _4       1.4  _i        1.4.7  _  1.4.7.10  - 


e 
&c. 

J57a;.  3.  Expand  (1  +x)~^  into  an  infinite  series. 

a     6a;8      6.1  la;3       6.11.16a;4 

5     5.10     5.10.15     5JlOll5.20     '      C* 

4.  Expand  (a3—  a;)""5  into  an  infinite  series. 

x      1.3z2     1.3.5z3     1.3.5.7ar4 


Ex.  5.  Expand  —====  into  an  infinite  series. 


c4      1.3c8     1.3.5cia  ,  1.3.5.7c18 

*         1          +~  + 


(274.)  The  binomial  theorem  may  be  employed  to  determine 
the  roots  of  surd  numbers. 

EXAMPLES. 

Ex.  I.  It  is  required  to  find  the  square  root  of  2. 
The  development  of  (a+b)*  has  been  given  in  Ex.  1,  page 

224.     If  we  make  a=l  and  6=1,  then  (a+b)*  becomes  (1+1)^ 
or  v/2;  and  the  terms  of  the  development  become 

1  __  1_       1.3        1.3.5         1.3.5.7 

2  2.4+2A6     2.4.6.8+2.4.6.8.10     ' 

which  therefore  expresses  the  square  root  of  2.     The  sum  of 


INVOLUTION    OF    BINOMIALS.  227 

this  series  is  1.41421.     As,  however,  the  series  converges  very 
slowly,  it  would  require  a  large  number  of  terms  to  give  the 
root  with  tolerable  accuracy.     The  following  example  affords 
a  better  illustration  of  the  utility  of  the  method. 
Ex.  2.  Required  the  square  root  of  101. 

.     Therefore 


Put  a=l  and  b=  JQ--  in  the  development  of  (a+b)2  on  page 
224,  and  we  shall  have 

S™1  =  W  (1+2loo""2.4.100»"l"2.4.6.1008"2.4.6.8.1004+f 

This  series  converges  so  rapidly  that  the  first  two  terms 
give  a  result  correct  to  three  decimal  places,  and  five  terms 
give  a  result  correct  to  ten  decimal  places. 

Thus,  the  value  of  the  first  term  is  1.00000000000 

"  second    "  +  .00500000000 

third       "  -  .00001250000 

"  fourth     "  +  .00000006250 

"  fifth         "  -   .00000000039 

Their  sum  is  1.00498756211. 

And  multiplying  by  10,  we  have 

VloT=  10.0498756211. 

Ex.  3.  It  is  required  to  convert  V9,  or  its  equal  (8+1)^,  into 
an  infinite  series,  and  find  its  value. 
Ans. 

1  1         __  5_  5.8  5.8.11 

~^~3.2*     3.6.2*     3.6.9.27     3.6.9.12.210     3.6.9.12.15.213     '         ' 

=2.08008. 
Ex.  4.  It  is  required  to  extract  the  cube  root  of  31. 

V31  =  3/27+4= 


--  -'- 

3.27     3.6.273"r3.6.9.278     3.6.9.  12.27*^ 

=3.14138. 


SECTION  XVII. 


EVOLUTION  OF  POLYNOMIALS. 

(275.)  Method  of  extracting  the  square  root  of  a  polynomial. 

In  order  to  discover  a  rule  for  extracting  the  square  root,  let 
us  consider  the  square  of  a+b,  which  is  d*+2ab+b*.  If  we 
write  the  terms  of  the  square  in  such  a  manner  that  the  pow- 
ers of  one  of  the  letters,  as  a,  may  go  on  continually  decreas- 
ing, the  first  term  will  be  the  square  of  the  first  term  of  the 
root ;  and  since  in  the  present  case  the  first  term  of  the  square 
is  a2,  the  first  term  of  the  root  must  be  a. 

Having  found  the  first  term  of  the  root,  we  must  consider 
the  rest  of  the  square,  namely,  2a&-}-&2,  to  see  how  we  can  de- 
rive from  it  the  second  term  of  the  root.  Now  this  remainder, 
2ab + 6s,  may  be  put  under  the  form  (2a+b)b  ;  whence  it  ap- 
pears that  we  shall  find  the  second  term  of  the  root  if  we  di- 
vide the  remainder  by  2a+b.  The  first  part  of  this  divisor, 
20,  is  double  of  the  first  term  already  determined ;  the  second 
part,  b,  is  yet  unknown,  and  it  is  necessary  at  present  to  leave 
its  place  empty.  Nevertheless,  we  may  commence  the  divis- 
ion, employing  only  the  term  2a;  but  as  soon  as  the  quotient 
is  found,  which,  in  the  present  case,  is  6,  we  must  put  it  in  the 
vacant  place,  and  thus  render  the  divisor  complete. 

The  whole  process,  therefore,  may  be  represented  as  fol- 
lows : 

d>+2ab+b*\a+b  =  the  root. 


2ab+b' 


2a+b  =  the  divisor. 


2db+V 
Hence  we  derive  the  following 


EVOLUTION    OF    POLYNOMIALS. 


RULE  FOB,  EXTRACTING  THE  SdUARE  ROOT  OF  A  POLYNOMIAL. 

Arrange  the  terms  according  to  the  powers  of  some  one  letter ; 
take  the  square  root  of  the  first  term  for  the  first  term  of  the  re- 
quired root,  and  subtract  its  square  from  the  given  polynomial 

Divide  the  first  term  of  the  remainder  by  double  the  root  al- 
ready found,  and  annex  the  result  both  to  the  root  and  the  divi- 
sor. Multiply  the  divisor  thus  increased  by  the  last  term  of  the 
root,  and  subtract  the  product  from  the  last  remainder.  Pro- 
ceed  in  the  same  manner  to  find  the  additional  terms  of  the  root. 

Ex.  1.  Required  the  square  root  of  a4— 2aax+3dlx*—2az*+x'. 
a*—2a*x+3a*x'*—2ax3+x*  I  a2— ax+x*  =  the  root. 


-2a3x+3a*x* 
-2a*x+  a 


2a?—ax  =  the  first  divisor. 


2a\x*-2ax3+x* 


2d*—2ax+x*  =  the  second  divisor. 


For  verification,  multiply  the  root  «2—  ax+x*  by  itself,  and 
we  shall  obtain  the  original  polynomial. 

Ex.  2.  Required  the  square  root  of  a2+2a&+2ac+6a+2&c+ca. 
Ex.  3.  Required  the  square  root  of  Wx*—  Wx9—  I2x*+5x*+ 


Ex.  4.  Required  the  square  root 


Ans.  2x*+2ax+4b\ 
Ex.  5.  Extract  the  square  root  of  15a46a+a6—  6a*b—  2Qa*b* 


Ex.  6.  Extract  the  square  root  of  8<z63+a4—  4asb+4b\ 

(276.)  Method  of  extracting  the  square  root  of  numbers. 

The  preceding  rule  is  applicable  to  the  extraction  of  the 
square  root  of  numbers.  For  every  number  may  be  regarded 
as  an  Algebraic  polynomial,  or  as  composed  of  a  certain  num- 
ber of  units,  tens,  hundreds,  &c.  Thus, 

529  is  equivalent  to  500+20+9. 

Also,  841  "  800+40+1. 

If,  then,  841  is  the  square  of  a  number  composed  of  tens 
and  units,  it  must  contain  the  square  of  the  tens,  plus  twice  the 


230  EVOLUTION    OF    POLYNOMIALS. 

product  of  the  tens  by  the  units,  plus  the  square  of  the  units. 
But  these  three  terms  are  blended  together  in  841,  and  hence 
the  peculiar  difficulty  in  determining  its  root.  The  following 
principles  will,  however,  enable  us  to  separate  these  terms, 
and  thus  detect  the  root. 

(277.)  I.  For  every  two  figures  of  the  square  there  will  be  one 
figure  in  the  root,  and  also  one  for  any  odd  figure. 

Thus,  the  square  of  1        is  1, 

"  10      is  100, 

100    is  10000, 
1000  is  1000000, 
&c.,  &c. 

The  smallest  number  consisting  of  two  figures  is  10,  and  its 
square  is  the  smallest  number  of  three  figures.  The  smallest 
number  of  three  figures  is  100,  and  its  square  is  the  smallest 
number  of  five  figures,  and  so  on.  Therefore,  the  square  root 
of  every  number  composed  of  one  or  two  figures  will  contain 
one  figure  ;  the  square  root  of  every  number  composed  of  three 
or  four  figures  will  contain  two  figures  ;  of  a  number  from  five 
to  six  figures  will  contain  three  figures ;  and  from  2n—  I  to  2n 
figures  must  contain  n  figures. 

Hence,  if  we  divide  the  number  into  periods  of  two  figures, 
proceeding  from  right  to  left,  the  number  of  figures  in  the  root 
will  be  equal  to  the  number  of  periods. 

(278.)  II.  The  first  figure  of  the  root  will  be  the  square  root 
of  the  greatest  square  number  contained  in  the  first  period  on 
the  left. 

For  the  square  of  tens  can  give  no  figure  in  the  first  right 
hand  period  ;  the  square  of  hundreds  can  give  no  figure  in  the 
first  two  periods  on  the  right ;  and  the  square  of  the  highest 
figure  in  the  root  can  give  no  figure  except  in  the  first  period 
on  the  left. 

Ex.  1.  Suppose  we  wish  to  find  the  square  root  of  529. 

The  square  of  23  or  20+3  is  20a+2.20.3+3a, 
or  400+120+9. 

Here  the  three  classes  of  terms  are  exhibited  distinct  from 
each  other,  and  we  might  extract  the  root  by  the  rule  of  Art. 
275.  But  observe  that  in  the  number  529,  since  the  square  of 
the  tens  can  not  give  a  figure  in  the  place  of  units  or  tens,  it 


EVOLUTION    OF    POLYNOMIALS.  231 

must  be  contained  in  the  first  period  5.  Now  this  period  con- 
tains not  only  the  square  of  the  tens,  but  also  a  part  of  the 
product  of  the  tens  by  the  units.  The  greatest  square  contain- 
ed in  5  is  4,  whose  root  is  2 ;  hence  2  must  be  the  number  of 
tens,  whose  square  is  400 ;  and  if  we  subtract  this  from  529, 
the  remainder  129  contains  twice  the  product  of  the  tens  by 
the  units,  plus  the  square  of  the  units.  If,  then,  we  divide  this 
remainder  by  twice  the  tens,  we  shall  obtain  the  units,  or  pos- 
sibly a  number  somewhat  too  large.  This  quotient  figure  can 
never  be  too  small,  but  it  may  be  too  large,  because  the  re- 
mainder 129,  besides  twice  the  product  of  the  tens  by  the  units, 
contains  the  square  of  the  units.  We  therefore  complete  the 
divisor  by  annexing  the  quotient  3  to  the  right  of  the  4,  and 
then  multiplying  by  3,  we  evidently  obtain  the  double  product 
of  the  tens  by  the  units,  plus  the  square  of  the  units.  The  en- 
tire operation  may  then  be  represented  as  follows : 

5'29|23=the  root 
4 

43|T29 
129. 

(279.)  Hence,  for  the  extraction  of  the  square  root  of  num- 
bers we  derive  the  following 

RULE. 

1.  Separate  the  given  number  into  periods  of  two  figures  each, 
beginning  at  the  right  hand. 

2.  Find  the  greatest  square  contained  in  the  left-hand  period ; 
its  root  is  the  first  figure  of  the  required  root.     Subtract  the 
square  from  the  first  period,  and  to  the  remainder  bring  down 
the  second  period  for  a  dividend. 

3.  Double  the  root  already  found  for  a  divisor,  and  find  how 
many  times  it  is  contained  in  the  dividend,  exclusive  of  its  right- 
hand  figure  ;  annex  the  result  both  to  the  root  and  the  divisor. 

4.  Multiply  the  divisor  thus  increased  by  the  last  figure  of  the 
root ;  subtract  the  product  from  the  dividend,  and  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend. 

5.  Double  the  whole  root  now  found  for  a  new  divisor,  and  con- 
tinue the  operation,  as  before,  until  the  periods  are  all  brought 
down. 

11 


23%  EVOLUTION    OF    POLYNOMIALS. 

Ex.  2.  Find  the  square  root  of  186624. 
The  operation  is  as  follows  ; 

18-66'24|432 
16 

83|  266 
249 


862|    17  24 
1724. 

Ex.  3.  Find  the  square  root  of  21086464. 
Ex.  4.  Find  the  square  root  of  88078225. 
(280.)  We  have  seen  that  the  root  of  a  fraction  is  equal  to 
the  root  of  its  numerator  divided  by  the  root  of  its  denominator. 

529  ^3 

Hence  the  square  root  of  —— ,  or  5.29,  is  — ,  or  2.3. 

The  square  root  of  ,nnnn,  or  18.6624,  is  — — ,  or  4.32. 

lUUUU  1UU 

That  is,  the  square  root  of  a  decimal  fraction  may  be  found  in 
the  same  manner  as  a  whole  number,  if  we  divide  it  into  periods 
commencing  with  the  decimal  point. 

Ex.  5.  Find  the  square  root  of  58.614336. 
Ex.  6.  Find  the  square  root  of  9.878449. 

Hence,  also,  if  the  square  root  of  a  number  can  not  be  found 
exactly,  we  may,  by  annexing  ciphers,  obtain  the  root  ap- 
proximately in  decimal  fractions. 

Ex.  7.  Find  the  square  root  of  2. 

Ans.  1.4142136  nearly. 

Ex.  8.  Find  the  square  root  of  3. 

Ex.  9.  Find  the  square  root  of  10. 

The  most  expeditious  method  of  extracting  roots  is  usually 
by  means  of  logarithms.  See  page  308. 

(281.)  Method  of  extracting  the  cube  root  of  a  polynomial. 

We  already  know  that  the  cube  of  a+b  is  «3+3a2&+3a6a+&3. 

If,  then,  the  cube  were  given,  and  we  were  required  to  find 
its  root,  it  might  be  done  by  the  following  method : 

When  the  terms  are  arranged  according  to  the  powers  of 
one  letter,  we  at  once  know,  from  the  first  term  a3,  that  a  must 
be  one  term  of  the  root.  If,  then,  we  subtract  its  cube  from 


EVOLUTION    OP    POLYNOMIALS.  233 

the  proposed  polynomial,  we  obtain  the  remainder  3a86+3a69 
+b9,  which  must  furnish  the  second  term  of  the  root. 
Now  this  remainder  may  be  put  under  the  form 


whence  it  appears  that  we  shall  find  the  second  term  of  the 
root,  if  we  divide  the  remainder  by  3a2+3a&+62.  But  as  this 
second  term  is  supposed  to  be  unknown,  the  divisor  can  not  be 
completed.  Nevertheless,  we  know  the  first  term  3a2,  that  is, 
thrice  the  square  of  the  first  term  already  found,  and  by  means 
of  this  we  can  find  the  other  part  b,  and  then  complete  the  di- 
visor before  we  perform  the  division.  For  this  purpose,  we 
must  add  to  3a2  thrice  the  product  of  the  two  terms,  or  Sab, 
and  the  square  of  the  second  term  of  the  root,  or  &2.  Hence 
we  derive  the  following 

RULE  FOR  EXTRACTING  THE  CUBE  ROOT  OF  A  POLYNOMIAL. 

(282.)  Arrange  the  terms  according  to  the  powers  of  some  one 
letter,  take  the  cube  root  of  the  first  term,  and  subtract  the  cube 
from  the  given  polynomial. 

Divide  the  first  term  of  the  remainder  by  three  times  the  square 
of  the  root  already  found,  the  quotient  will  be  the  second  term  of 
the  root. 

Complete  the  divisor  by  adding  to  it  three  times  the  product  of 
the  two  terms  of  the  root,  and  the  square  of  the  second  term. 

Multiply  the  divisor  thus  increased  by  the  last  term  of  the  root, 
and  subtract  the  product  from  the  last  remainder.  Proceed  in 
the  same  manner  to  find  the  additional  terms  of  the  root. 

Ex.  1.  Extract  the  cube  root  of  a3+12a2+48a+64. 
a3+12a2+48a+64   a+4  =  the  root. 


3a2+12«+16  =  the  divisor. 


Having  found  the  first  term  of  the  root  a,  and  subtracted  its 
cube,  we  divide  the  first  term  of  the  remainder,  12a2,  by  three 
times  the  square  of  a,  that  is,  3«2,  and  we  obtain  4  for  the  sec- 
ond term  of  the  root.  We  then  complete  the  divisor  by  add- 
ing to  it  three  times  the  product  of  the  two  terms  of  the  root, 
which  is  120,  together  with  the  square  of  the  last  term,  4, 


234  EVOLUTION    OF    POLYNOMIALS. 

which  is  16.  Multiplying,  then,  the  complete  divisor  by  4, 
and  subtracting  the  product  from  the  last  remainder,  nothing 
is  left.  Hence  the  required  cube  root  is  0+4. 

This  result  may  be  easily  verified  by  multiplication. 

Ex.  2.  Extract  the  cube  root  of  a6—  6a6+15a4-20<z8+15a* 


a'-6a6+15a4-20<z8+15a2-6a-}-l  |<za-2<z+l  =  the  root. 


-6a6+15a4-20a8 


3a4—  6a3+4a3  =  the  first  divisor. 


3a4— 
3a4-12as+1503-6a+l 


3a4— 


the  second  divisor. 


We  may  dispense  with  forming  the  complete  divisor  accord- 
ing to  the  rule,  if,  each  time  that  we  find  a  new  term  of  the 
root,  we  raise  the  entire  root  to  the  third  power,  and  subtract 
the  cube  from  the  given  polynomial. 

Ex.  3.  Required  the  cube  root  of  6x*-40x3+x'+96x—64. 

Ex.  4.  Required  the  cube  root  of  18a;4+36:c3-f-24x+8+32:r3 
+x°+6x6. 

Ex.  5.  Required  the  cube  root  of  3&5+&6— 5fe8—  l+3fc. 

(283.)  Method  of  extracting  the  cube  root  of  numbers. 

The  preceding  rule  is  applicable  to  the  extraction  of  the 
cube  root  of  numbers ;  but  a  difficulty  in  applying  it  arises 
from  the  fact  that  the  terms  of  the  power  are  all  blended  to- 
gether in  the  given  number.  They  may,  however,  be  separated 
by  attending  to  the  following  principles : 

I.  For  every  three  figures  of  the  cube  there  will  be  one  figure 
in  the  root,  and  also  one  for  any  additional  figure  or  figures. 

Thus,  the  cube  of  1        is  1, 

10      is  1000, 
100    is  1000000, 
1000  is  1000000000, 
&c.,  &c. 

Hence  we  see  that  the  cube  root  of  a  number  consisting  of 
from  1  to  3  figures  will  contain  one  figure  ;  of  a  number  from 
4  to  6  figures  will  contain  two  figures ;  from  7  to  9  figures  will 
contain  three ;  and  from  3w— 2  to  3n  figures  must  contain  n 


EVOLUTION    OF    POLYNOMIALS.  233 

figures.  Hence,  if  we  divide  the  number  into  periods  of  three 
figures,  proceeding  from  right  to  left,  the  number  of  figures  in 
the  root  will  be  equal  to  the  number  of  periods. 

II.  The  first  figure  of  the  root  will  be  the  cube  root  of  the  great- 
est cube  number  contained  in  the  first  period  on  the  left. 

Ex.  1.  Suppose  we  wish  to  find  the  cube  root  of  12167. 

The  cube  of  23  or  20+3  is  203+3.20a.3+3.20.33+33, 
or  8000+3600+540+27. 

Here  the  four  classes  of  terms  are  exhibited  distinct  from 
each  other,  and  the  rule  of  Art.  282  might  be  easily  applied. 
But  observe  that  in  the  number  12167,  since  the  cube  of  the 
tens  can  not  give  a  figure  in  the  first  three  places,  it  must  be 
contained  in  the  first  period  12.  The  greatest  cube  contained 
in  this  is  8,  the  root  of  which  is  2.  Hence  2  must  be  the  num- 
ber of  tens  whose  cube  is  8000  ;  and  the  remainder  4167  con- 
tains three  times  the  product  of  the  square  of  the  tens  by  the  units , 
plus  three  times  the  product  of  the  tens  by  the  square  of  the  unitsf 
plus  the  cube  of  the  units. 

If,  then,  we  divide  this  remainder  by  three  times  the  square 
of  the  tens,  we  shall  obtain  the  units,  or  possibly  a  number  too 
large,  because  the  divisor  is  too  small.  We  therefore  com- 
plete the  divisor  by  adding  to  it  three  times  the  product  of  the 
tens  by  the  units,  plus  the  square  of  the  units.  The  entire 
operation  is  then  as  follows : 

12-167|23=the  root. 

8 


20aX3        =1200 

20  X3X3  =    180 

32=        9 


4167 


4  167 


complete  divisor  =1389 

(284.)  Hence,  for  the  extraction  of  the  cube  root  of  numbers, 
we  derive  the  following 


RULE. 


1 .  Separate  the  given  number  into  periods  of  three  figures 
each,  beginning  at  the  right  hand. 

2.  Find  the  greatest  cube  contained  in  the  left-hand  period ; 
its  root  is  the  first  figure  of  the  required  root.     Subtract  the  cube 


236  EVOLUTION  OF  POLYNOMIALS. 

from  the  first  period,  and  to  the  remainder  bring  down  the  sec- 
ond  period  for  a  dividend. 

3.  Take  three  hundred  times  the  square  of  the  root  already 
found  for  a  trial  divisor;  find  how  many  times  it  is  contained 
in  the  dividend,  and  place  the  quotient  for  a  second  figure  of  the 
root. 

4.  Complete  the  divisor  by  adding  to  it  thirty  times  the  prod- 
uct of  the  two  figures  of  the  root,  and  the  square  of  the  second 
figure. 

5.  Multiply  the  divisor  thus  increased  by  the  last  figure  of 
the  root ;  subtract  the  product  from  the  dividend,  and  to  the  re- 
mainder bring  down  the  next  period  for  a  new  dividend. 

6.  Take  three  hundred  times  the  square  of  the  whole  root  now 
found  for  a  new  trial  divisor,  and  continue  the  operation  as  be- 
fore until  all  the  periods  are  brought  down. 

It  will  be  observed  that  three  times  the  square  of  the  tens, 
when  their  local  value  is  regarded,  is  the  same  as  three  hun- 
dred times  the  square  of  this  digit,  not  regarding  its  local 
value. 

Ex.  2.  Find  the  cube  root  of  20796875. 

Ex.  3.  Find  the  cube  root  of  2509911279. 

Ex.  4.  Find  the  cube  root  of  895562584U9. 

The  same  method  is  applicable  to  the  extraction  of  the  cube 
root  of  fractions,  and  also  of  imperfect  powers. 

Ex.  5.  Find  the  cube  root  of  604.422796375. 

Ex.  6.  Find  the  cube  root  of  4. 

Ex.  7.  Find  the  cube  root  of  11. 

(285.)  Method  of  extracting  any  root  of  a  polynomial. 

We  already  know  that  the  nth  power  of  a+b  is  an+nan~1b+ 
other  terms.  The  first  term  of  the  root  is,  therefore,  the  rath 
root  of  the  first  term  of  the  polynomial.  Also,  the  second  term 
of  the  root  may  be  found  by  dividing  the  second  term  of  the 
polynomial  by  nan~l ;  that  is,  the  first  term  of  the  root  raised 
to  the  next  inferior  power,  and  multiplied  by  the  exponent  of 
the  given  power.  Hence  we  deduce  the  following 

RULE  FOB,  EXTRACTING  ANY  ROOT  OF  A  POLYNOMIAL. 

Arrange  the  terms  according  to  the  powers  of  one  of  the  letters, 


EVOLUTION    OF    POLYNOMIALS.  237 

and  take  the  nth  root  of  the  first  term  for  the  first  term  of  the  re- 
quired root. 

Subtract  its  power  from  the  given  polynomial,  and  divide  the 
first  term  of  the  remainder  by  n  times  the  (n—  1)  power  of  this 
root ;  the  quotient  will  be  the  second  term  of  the  root. 

Subtract  the  nth  power  of  the  terms  already  found  from  the 
given  quantity,  and,  using  the  same  divisor,  proceed  in  like  man- 
ner to  find  the  remaining  terms  of  the  root. 

Ex.  1.  Required  the  fourth  root  of  16a4— 96a3#-H216aV— 
21Gax3+8lx\ 

16a4-96a3o;+216aV-216az3+81a;4|2<z— 3x  =  the  root. 
16a4 

—  96a3x\32a*  =  the  divisor. 


Here  we  take  the  fourth  root  of  16«*,  which  is  2a,  for  the 
first  term  of  the  required  root  ;  subtract  its  fourth  power,  and 
bring  down  the  first  term  of  the  remainder  —  96a3x.  For  a 
divisor,  we  raise  the  first  term  of  the  root  to  the  third  power, 
and  multiply  it  by  4,  making  32as.  Dividing,  we  obtain  —  3x 
for  the  second  term  of  the  root.  The  quantity  2a—  3x  being 
raised  to  the  fourth  power,  is  found  to  be  equal  to  the  proposed 
polynomial. 

Ex.  2.  Required  the  fifth  root  of  80x*+32x*-8Qx*—  40^+ 
lOx-1. 

AnS.    2X-1. 

Ex.  3.   Required  the  fourth  root  of  336xf'+81x*-2l6x'- 


Ans.  3x'—2x—2. 

(286.)  Method  of  extracting  any  root  of  numbers. 

It  is  easy  to  apply  the  preceding  Rule  to  the  extraction  of 
any  root  of  numbers.  For  a  reason  similar  to  that  given  for 
the  square  and  cube  roots,  we  must  first  divide  the  number  into 
periods  of  n  figures  each.  Then  the  first  figure  of  the  root 
will  be  the  nth  root  of  the  greatest  nth  power  contained  in  the 
first  period  on  the  left.  If  we  subtract  its  power  from  the- 
given  number,  and  divide  the  remainder  by  n  times  the  (n  —  1) 
power  of  the  first  figure,  regarding  its  local  value,  the  quotient 
will  be  the  second  figure  of  the  root,  or,  possibly,  something 
too  large.  The  result  may  be  verified  by  raising  the  whole 


238  EVOLUTION  OF  POLYNOMIALS. 

root  now  found  to  the  nth  power  ;  and  if  ther*  are  other  figures, 
they  may  be  found  in  the  same  manner. 
Ex.  1.  Find  the  fifth  root  of  33554432. 
335'54432|32 
243 
5.34=405|  925 

326=335  54432. 

Ex.  2.  Find  the  fifth  root  of  4984209207. 
Ex.  3.  Find  the  fifth  root  of  10. 

(287.)  When  the  index  of  the  root  to  be  extracted  is  a  mul- 
tiple of  two  or  more  numbers,  we  may  obtain  the  root  required 
by  the  successive  extraction  of  simpler  roots,  Art.  159. 

For  example,  we  may  obtain  the  fourth  root  by  extracting 
the  square  root  twice  successively  ;  for  the  square  root  of  a*  is 
a3,  and  the  square  root  of  a3  is  a. 

The  eighth  root  may  be  obtained  by  extracting  the  square 
root  three  times  successively  ;  for  the  square  root  of  a3  is  a4, 
that  of  a*  is  a2,  and  that  of  a3  is  a. 

In  the  same  manner,  the  sixteenth  root  may  be  obtained  by 
extracting  the  square  root  four  times  successively,  and  so  on. 

The  sixth  root  may  be  found  by  extracting  the  square  root, 
and  afterward  the  cube  root  ;  for  the  square  root  of  a6  is  a8, 
and  the  cube  root  of  a8  is  a.  We  may  also  take,  first,  the  cube 
root,  which  gives  a3,  and  afterward  the  square  root,  which 
gives  a,  as  before.  It  is,  however,  best  to  extract  the  roots  of 
the  lowest  degree  first,  because  the  operation  is  less  laborious. 

In  general,  the  rrmtJi  root  of  a  number  is  equal  to  the  nth  root 
of  the  mth  root  of  this  number.  That  is, 


For,  raising  each  member  of  this  equation  to  the  nth  power, 
we  have 


Ex.  1.  Find  the  fourth  root 
Ex.  2.  Find  the  sixth  root  of  6a6b+15a'b9+a°+20a3b*+15a9b4 
+b'+6ab6. 

Ex.  3.  Find  the  eighth  root  of  1024z7y+1792zy+256:c''-f- 
+  1792a? 


EVOLUTION   OF    POLYNOMIALS.  239 


EXTRACTION  OF  THE  SQUARE  ROOT  OF  A  QUANTITY  OF  THE  FORM 


(288.)  Binomials  of  this  class  require  particular  attention, 
because  they  frequently  occur  in  the  solution  of  equations  of 
the  fourth  degree,  such  as  are  treated  of  in  Art.  184.  Thus 
the  equation 


gives  us  a:a=7±4x/3. 

Hence,  in  order  to  find  the  value  of  x,  we  must  extract  the 
square  root  of  the  binomial  7=1=4^3. 

In  order  to  show  that  the  square  root  of  such  an  expression 
may  sometimes  be  extracted,  take  the  binomial 


and  find  its  square. 


Therefore,  the  square  root  of  7±4v/3  is  2±v/3. 

The  square  root  of  an  expression  of  the  form  a±  </b  may, 
therefore,  sometimes  be  extracted,  and  it  is  required  to  deter- 
mine a  general  method  for  this  purpose  whenever  it  is  prac- 
ticable. 

THEOREM  I. 

(289.)  The  sum  or  difference  of  two  surds  can  not  be  equal  to 
a  rational  quantity. 

For,  if  possible,  let  ^/a±^/b=c,  where  c  denotes  a  rational 
quantity,  and  v/a,  ^/b  denote  surd  quantities. 

By  transposing  ^/b  and  squaring  both  sides,  we  obtain  a= 
whence,  by  transposition  and  division,  we  have 


The  second  member  of  the  equation  contains  only  rational 
quantities,  while  ^/b  was  supposed  to  be  irrational;  that  is, 
we  find  an  irrational  quantity  equal  to  a  rational  one,  which 
is  absurd.  Hence  the  sum  or  difference  of  two  surds  can  not 
be  equal  to  a  rational  quantity. 

11* 


240  EVOLUTION    OF    POLYNOMIALS. 

THEOREM  II. 

In  every  equation  of  the  form 


the  rational  parts  on  the  opposite  sides  are  equal  to  each  other, 
and  also  the  irrational  parts. 

For  if  x  is  not  equal  to  <z,  let  it  be  equal  to  a±z. 

Then 


or  z=^b-^y; 

that  is,  a  rational  quantity  is  equal  to  the  difference  of  two 
surds,  which,  by  the  last  Theorem,  is  impossible.  Therefore, 
x—a,  and,  consequently,  Vy=  ^b. 

THEOREM  III. 

If  Va+  Vb  is  equal  to  x+  x/y, 

then  will  Va—  Vb  be  equal  to  x—  </y. 

For,  by  involution,  a+^b=x*-t-2x,/y+y. 

But,  by  the  last  Theorem,  a=x*+y, 

and  </b 

Subtracting,  we  obtain       a—^b 

Therefore,  by  evolution,  Va—  Vb—x—  </y. 

(290.)   To  find  an  expression  for  the  square  root  of  a 

Let  us  assume          Va+  Vb=p-\-q  (1), 

where  p  and  q  may  be  both  radicals,  or  one  rational  and  the 
other  a  radical,  but  jo2  and  cf  are  required  to  be  rational. 

Then,  by  the  last  Theorem,  ^  . 

Va—  Vb=p—q  (2). 

Multiplying  these  equations  together,  we  obtain 

Va*—b=p*—q*  (3),  a  rational  quantity. 

Hence  we  see  that,  in  order  that  ^Ja+  Vb  may  be  expressed 
by  the  sum  of  two  radicals,  or  one  rational  term  and  the  other 
a  radical,  the  expression  a3—  b  must  be  a  perfect  square. 

Let,  then,  aa—  b  be  a  perfect  square,  and  put  Va'—b=c; 
equation  (3)  will  thus  become 


EVOLUTION    OF    POLYNOMIALS.  241 

Squaring  equations  (1)  and  (2),  we  obtain 


Adding  these  two  equations,  we  obtain 

But  we  have  already  found 

p'-q'=c. 

Hence  2/?2=a+c, 

and  2<7a=a— c. 

From  which  we  obtain 

ta+c 

p  =  : 


/a— c 
and  q-  tV~~2~- 

Therefore, 


.  -  -  ,/     a+c        /ov 

Va-Vb,orp-q=  fc  (y  —  -  \J  —J  . 

(291.)  Hence,  to  extract  the  square  root  of  a  binomial  of  the 
form  a±  -x/fe,  we  have  the  following 


RULE. 


From  the  square  of  the  rational  part  (a2),  take  the  square  of 
the  irrational  part  (b)  ;  extract  the  square  root  of  the  remainder  ', 
and,  calling  that  root  c,  the  required  root  will  be 


a+c          a—c 


Ex.  1.  Required  the  square  root  of  4+2  -v/3. 

Here  a=4,  and  ^6=2  x/3;  therefore,  ai-b=c^=  16-  12=4; 
or  c=2.  Hence,  by  the  above  formula,  the  required  root  will 
be 


Verification. 

The  square  of  v/3+1  is  3+2^3+1=4+2^3. 

Q 


242  EVOLUTION    OF    POLYNOMIALS. 

Ex.  2.  Required  the  square  root  of  11+6^/2. 

Here  a=ll,  and  v/6=6v/2;  therefore,  6=36X2=72;  and 
a8—  6=49=ca.  Hence  c=7,  and  we  find  the  square  root  of 
ll+6v/2  is  x/9+^/2,  or  3+^2.  Ans. 

Ex.  3.  Required  the  square  root  of  11—  2^30. 

Ans.  ^6—  x/5. 

Ex.  4.  Required  the  square  root  of  2+v/3. 

Ans.  ^f+v/i. 

(292.)  This  method  is  applicable  even  when  the  binomial 
contains  imaginary  quantities. 

Ex.  5.  Required  the  square  root  of  1+4  V  —  3. 
Here  a=l,  and  </b=4V—  3;  hence  &=—  48,andaa—  6=49; 
therefore,  c=7.     The  required  square  root  is  </4+  V—  3=2+ 


j&c.  6.  Required  the  square  root  of  —  i+|Vf  —  3. 

Ans. 

j&£.  7.  Required  the  square  root  of  2V  —  1. 

Here  we  put  a=0  ;  hence  c=2,  and  the  required  root  is 

1+V^T, 

which  may  be  easily  verified. 

Ex.  S.  Required  the  value  of  the  expression 


Ex.  9.  Required  the  value  of  the  expression 

Ans.  6. 


SECTION  XVIII. 


INFINITE   SERIES. 

(293.)  An  infinite  series  is  an  infinite  number  of  terms,  each 
of  which  is  derived  from  the  preceding  term  or  terms  accord- 
ing to  some  law. 

As  l+i+i+t+rV+.&c., 

or  i-i.+|-_V+_V_,&c. 

These  are  examples  of  geometrical  progressions,  in  the  first 
of  which  the  ratio  is  |,  and  in  the  second  it  is  —  J. 

Infinite  series  may  arise  from  the  common  operations  of  di- 
vision, the  extraction  of  roots,  and  other  processes  of  calcula- 
tion, as  will  be  seen  hereafter. 

A  converging  series  is  one  in  which  the  sum  of  any  number 
of  its  terms  is  finite,  as  in  the  examples  just  given. 

A  diverging  series  is  one  in  which  the  sum  of  its  terms  is  not 
finite  ;  as, 

1+2+3+4+5+6+7,  &c. 

An  ascending  series  is  one  in  which  the  exponents  of  the  un- 
known quantity  continually  increase  ;  as, 

ax+bx*+cx'+dx'+ex*+,  &c. 

A  descending  series  is  one  in  which  the  exponents  of  the  un- 
known quantity  continually  decrease  ;  as, 

+,  &c. 


PROBLEM  I. 

(294.)  Any  series  being  given,  to  find  its  several  orders  of 
differences. 


244  INFINITE    SERIES. 


RULE. 

1.  Take  the  first  term  from  the  second,  the  second  from  the 
third,  the  third  from  the  fourth,  fyc. ;  and  the  remainders  will 
form  a  new  series,  called  the  FIRST  ORDER  OF  DIFFERENCES. 

2.  Take  the  first  term  of  this  last  series  from  the  second,  the 
second  from  the  third,  fyc. ;  and  the  remainders  will  form  a  third 
series,  called  the  SECOND  ORDER  OF  DIFFERENCES. 

3.  Proceed  in  like  manner  for  the  third,  fourth,  fyc.,  orders 
of  differences,  and  so  on  till  they  terminate,  or  are  carried  as  far 
as  may  be  thought  necessary. 

Ex.  1.   Required  the  several  orders  of  differences  of  the 
series  of  squares, 

1     49     16     25     36     49,  &c. 
357       9      11     13          first  differences. 
22222          second  differences. 
00       0      0  third  differences. 

Ex.  2.   Required  the  several  orders  of  differences  of  the 
series  of  cubes, 

1     8     27     64     125     216,  &c. 

7    19     37     61       91  first  differences. 

12     18     24     30  second  differences. 

666  third  differences. 

0       0  fourth  differences. 

Ex.  3.  Required  the  several  orders  of  differences  of  the 
series  of  fourth  powers, 

1,  16,  81,  256,  625,  1296,  &c. 

Ex.  4.   Required  the  several  orders  of  differences  of  the 
series  of  fifth  powers, 

1,  32,  243,  1024,  3125,  7776,  16807,  &c. 

Ex.  5.   Required  the  several  orders  of  differences  of  the 
series  of  numbers, 

1,  3,  6,  10,  15,  21,  &c. 

PROBLEM  II. 

(295.)  To  find  the  nth  term  of  the  series 
a,  b,  c,  d,  e,  &c. 


INFINITE    SERIES.  245 

Take  the  proposed  series,  and  subtract  each  term  from  the 
next  succeeding  one ;  we  shall  thus  obtain  for  the  first  order 
of  differences, 

b—a,  c—b,  d—c,  e—d,  &c. 

Again,  subtracting  each  term  of  this  series  from  the  next 
succeeding  term,  we  find  for  the  second  order  of  differences, 

c—2b+a,  d—2c+b,  e—2d+c,  &c. 

Subtracting,  again,  each  term  of  the  preceding  series  from 
its  next  succeeding  term,  we  find  the  third  order  of  differences, 

d-3c+3b-a,  e—3d+3c-b,  &c. 
Subtracting  again,  we  find  for  the  fourth  order  of  differences, 

Let  D',  D",  D'",  D"",  &c.,  represent  the  first  terms  of  the 
several  orders  of  differences. 

Then, 

D'    =b—a;  whence  b=a+  D', 

D"  =c-2b+a;  "       c=a+2D'+  D", 

T>'"=d-3c+3b-a;  "       d=a+3V'+3V"+  D"', 

V""=e-4d+6c-4b+a;      "        e=<z+4D'+6D"+4D'"+D"", 
&c.,  &c. 

The  coefficients  of  the  value  of  c,  the  third  term  of  the  pro- 
posed series,  are  1,  2,  1,  which  are  the  coefficients  of  the  sec- 
ond power  of  a  binomial ;  the  coefficients  of  the  value  of  d,  the 
fourth  term,  are  1,  3,  3,  1,  which  are  the  coefficients  of  the 
third  power  of  a  binomial,  and  so  on.  Hence  we  infer  that 
the  coefficients  of  the  nth  term  of  the  series  are  the  coefficients 
of  the  (n—  1)  power  of  a  binomial.  Therefore,  the  nth  term 
of  the  series  will  be 


.  &c. 


Ex.  1.  Required  the  twelfth  term  of  the  series 

2,  6,  12,  20,  30,  &c. 

The  first  order  of  differences  is  468  10,  &c. 
The  second  order  of  differences  is  2  2  2,  &c. 
The  third  order  of  differences  is  0  0. 

Here  D'=4,  D"=2,  and  D'"=0.    Also,  a=2,  and  n=l2. 


246  INFINITE    SERIES. 

Hence  a+(n-l)V'+(n~l)  (^~2)D//=2+llD'+55D"= 

2+44+110=156  =  the  twelfth  term. 
Ex.  2.  Required  the  twentieth  term  of  the  series 

1,  3,  6,  10,  15,  21,  &c. 

Here  <z=l,  D'=2,  D"=l,  and  n=20. 

Therefore,  the  20th  term  =  l  +  19D'+17lD"=l+38+171 

=210,  Ans. 
Ex.  3.  Required  the  thirteenth  term  of  the  series 

1,  5,  14,  30,  55,  91,  &c. 

Ex.  4.  Required  the  fifteenth  term  of  the  series 
1,  4,  9,  16,  25,  36,  &c. 

Ans.  225. 

Ex.  5.  Required  the  twentieth  term  of  the  series 
1,  8,  27,  64,  125,  &c. 

PROBLEM  III. 

(296.)   To  find  the  sum  ofn  terms  of  the  series 

a,  b,  c,  d,  e,  &c. 
Assume  the  series 

0,  a,  a+b,  a+b+c,  a+b+c+d,  &c. 
Subtracting  each  term  from  the  next  succeeding,  we  obtain 

a,  b,  c,  d,  e,  &c., 

which  is  the  series  whose  sum  it  is  proposed  to  find.  Hence 
the  sum  of  n  terms  of  the  proposed  series  is  the  (?i+l)th  term 
of  the  assumed  series  ;  and  the  rath  order  of  differences  in  the 
first  series  is  the  (?i+l)th  order  in  the  other  series.  If,  there- 
fore, in  the  formula  of  the  preceding  Problem,  we  substitute 

0  for  a, 
n+ 1  for  n, 
a  for  D', 
D'  for  D", 

&c., 
we  shall  have 

na  I  "fo-^D'  .  "fr"1)  (n~2V  i  "fo-1)  (n~V  (n-Vp'»  I 
2  2.3  2.3.4 

&c., 

which  is  the  sum  of  n  terms  of  the  proposed  series. 


INFINITE   SERIES.  247 

Ex.  1.  Required  the  sum  of  n  terms  of  the  series 

1,  2,  3,  4,  5,  6,  &c. 
Here  a=I,  D'=l,  D"=0. 

i  71(71 
Therefore,  na+ 


g 

the  sum  of  TI  terms,  the  same  as  found  in  Art.  239. 
Ex.  2.  Required  the  sum  of  n  terms  of  the  series 

I3,  22,  3a,  4a,  52,  &c. 
Here  a=l,  D'=3,  D"=2. 

Therefore  the  general  formula  reduces  to 

371(71-1)       27l(7l-l)  (71  —  2) 

~~~~  ~  ' 


71(71+1)  (271  +  1) 

—  jp-      —  ,  the  sum  required. 

j&c.  3.  Required  the  sum  of  n  terms  of  the  series 

I3,  2s,  33,  43,  53,  63,  &c. 
Here  a=l,  D'=7,  D/;=12,  D^'^B. 


?.  4.  Required  the  sum  of  n  terms  of  the  series 
1,  3,  6,  10,  15,  &c. 

7l(7l+l)  (71  +  2) 


Ex.  5.  Required  the  sum  of  n  terms  of  the  series 
1,  4,  10,  20,  35,  &c. 

n(n+l)  (n+2)  (n+3) 
2.3.4 

PROBLEM  IV. 

(297.)  A  series  of  equidistant  terms,  a,  b,  c,  d,  e,  fyc.t  being 
given,  to  find  any  intermediate  term  by  interpolation. 

This  is  essentially  the  same  as  Problem  II.  For  convenience, 
let  us  put  x  to  represent  the  distance  of  the  required  term  from 
the  first  term  of  the  series  a,  in  which  case  x=n—  1,  and  we 
shall  have 


248  INFINITE   SERIES. 


'+.  &c. 


Ex.  1.  Given  the  square  root  of  94,  equal  to  9.69536  ; 
«  95,        "        9.74679; 

"  "  96,        "        9.79796, 

to  find  the  square  root  of  94.25. 

Here  the  first  differences  are  +.05143,  +.05117, 
and  the  second  difference  is  —.00026; 

that  is,  D'=+.05143,  D"=-.00026. 

But  z=a+JD'-33¥D". 

Hence  the  square  root  of  94.25  is 

9.69536+.01286+.00002, 
or  9.70824.  Ans. 

Ex.  2.  Given  the  square  root  of  160,  equal  to  12.64911  ; 
"  "  162,        "        12.72792; 

"  "  164,         "         12.80625, 

to  find  the  square  root  of  161. 

Here  the  interval  between  the  given  numbers  is  2  ;  the  dis- 
tance of  the  required  term  from  the  first  term  is  1  ;  and,  since 
the  interval  of  the  given  numbers  is  always  to  be  reckoned  as 
unity,  we  have  x=^. 

Also,  D'=  +.07881,  D"=  -.00048. 

And  z^a+JD'—  |D". 

Therefore  the  square  root  of  161  is 

12.6491  1  +.03941  +.00006, 
or  12.68858.  Ans. 

Ex.  3.  Given  the  cube  root  of  60,  equal  to  3.91487  ; 

"  «  62,         "        3.95789; 

"  "  64,         "        4.00000; 

"  "  66,         "        4.04124, 

to  find  the  cube  root  of  61. 

Ans.  3.93650. 

Ex.  4.  Given  the  fourth  root  of  625,  equal  to  5.000000  ; 
"  «  628,         "       5.005988; 

"  "  631,         "       5.011956; 

634,         "       5.017903, 
to  find  the  fourth  root  of  627. 

Here  z=f.     Therefore,  z=a+fD'-iD". 

Ans.  5.003994. 


INFINITE    SERIES.  249 

Ex.  5.  Given  the  square  root  of  70,  equal  to  8.36660 ; 

"  "  74,         "       8.60233; 

78,         "        8.83176; 

82,         "        9.05539, 

to  find  the  square  root  of  71. 

Ans.  8.42615. 

(298.)  Fractions  expanded  into  infinite  series. 
When  the  dividend  is  not  exactly  divisible  by  the  divisor, 
the  quotient  may  be  expressed  by  a  fraction.     Thus,  if  it  is 

required  to  divide  1  by  1—  a,  we  obtain  the  fraction . 

We  may,  however,  commence  the  division  according  to  the 
usual  method ;  thus, 

I  — a 

-«3+a4+,  &c.,  =  the  quotient. 


a4 

Hence  —  —  =  l+a+«a+a8+«44-a6+,  &c.,  to  infinity. 
Suppose  a=^,  we  shall  then  have 

=  —  T—  2»  which  will  be  equal  to  the  series 


Suppose  a=J,  we  shall  then  have 

-=       1  =  |,  which  will  be  equal  to  the  series 


1 


Ex.  2.  Resolve  -    -  into  an  infinite  series. 
1+a 

Ans.  l-a+aa—  a8+<z4-a'+,  &c. 
Suppose  a—  -J-,  we  shall  then  have 


250  INFINITE    SERIES. 

1 


=|-,  which  will  be  equal  to  the  series 


Ex.  3.  Resolve  the  fraction  — — r  into  an  infinite  series. 

a+b 

c     be    Vc    b*c 

Ans. H — = r+>  otc. 

a     a      a       a 

Ex.  4.  Resolve  ^— —  into  an  infinite  series. 
b+x 

Ex.  5.  Resolve into  an  infinite  series. 

l—x 

We  may  proceed  in  the  same  manner  when  there  are  more 
than  two  terms  in  the  divisor. 

Ex.  6.  Resolve ; — 5  into  an  infinite  series. 


Ans.  I+a—a*— a4+«6+,  &c. 
Ex.  7.  Resolve  -: — • — rr  into  an  infinite  series. 


(299.)  Infinite  series  obtained  by  extracting  the  square  root. 

In  Art.  272,  Va+b  has  been  expanded  into  an  infinite  series 
by  the  Binomial  Theorem.  It  was  also  remarked  that  the 
same  result  might  have  been  obtained  by  extracting  the  square 
root  according  to  the  usual  rule,  Art.  275.  The  operation  will 
proceed  as  follows  : 


a+b 


a*-\ r jH 5—,  &c.,  =  the  square  root  of  a+b. 


2a* 


2a2H — r,  first  divisor. 
2 


*     b      b* 
r  H — Y 3 ,  second  divisor. 

a2     Sa2 


___ 

4a     8aa  64a* 

,  j^_  JL 

8a9  64a8* 


INFINITE    SERIES.  251 

This  result  is  the  same  as  that  obtained  in  Art.  272. 
Ex.  2.  Extract  the  square  root  of  1— #. 

x    x*    x3     5x* 
Ans.  !_________,  &c. 

Ex.  3.  Extract  the  square  root  of  cf+b. 
Ex.  4.  Extract  the  square  root  of  a*— b. 

METHOD  OF  UNKNOWN  COEFFICIENTS. 

(300.)  The  method  of  unknown  coefficients  is  a  method  of 
developing  algebraic  expressions  into  series,  by  assuming  a 
series  having  unknown  coefficients,  and  afterward  finding  the 
value  of  these  coefficients.  This  method  is  founded  on  the  fol- 
lowing 

THEOREM. 

If  an  equation  of  the  form 

A+Bz+Oc2+Da;3+,  &c.,  =A'+R'x+C'x*+V'x3+,  &c., 
must  be  verified  by  any  value  given  to  x,  the  terms  involving  the 
same  powers  in  the  two  members  are  respectively  equal. 

For,  since  this  equation  must  be  verified  for  every  value  of 
x,  it  must  be  verified  when  x=0.  But,  upon  this  supposition, 
all  the  terms  vanish  except  two,  and  we  have 

A=A'. 
Suppressing  these  two  equal  terms,  we  have 

Bz+Oc2+D.r'+,  &c.,  -=B^+CV+DV+,  &c. 
Dividing  every  term  by  x,  we  obtain 

B+Cx+Vx>+,  &c.,  =B/+C'z+D'or2+,  &c. 
Since  this  equation  must  be  verified  for  every  value  of  #,  it 
must  be  verified  when  x=0.     But,  upon  this  supposition, 

B=B'. 

In  the  same  manner,  we  can  prove  that 
C=C', 
D=D',  &c. 

(301.)  Let  it  be  proposed  to  develop  the  expression  — 

into  a  series  arranged  according  to  the  powers  of  x.  It  is 
plain  that  this  development  is  possible,  for  we  may  divide  the 
numerator  by  the  denominator,  as  explained  in  Art.  298. 


252  INFINITE    SERIES. 

Let  us,  then,  assume 

,  &c., 


where  the  coefficients  A,  B,  C,  D  are  supposed  to  be  independ- 
ent of  xt  but  dependent  on  the  known  terms  of  the  fraction. 

In  order  to  obtain  the  values  of  these  coefficients,  let  us  mul- 
tiply both  members  of  the  above  equation  by  the  denominator 
1  +x,  and  we  shall  have 

l-a;=A+(A+B>+(B+CX+(C+D)a;8+(D+EX+,  &c. 

But,  according  to  the  preceding  Theorem,  the  terms  involving 
the  same  powers  of  a;  in  the  two  members  of  the  equation  must 
be  equal  to  each  other. 
Therefore,  A=     1, 

A+B  =  -1;  hence  B  =  -2. 
=     0;      «      C  =  +2. 
0;       "      D=-2. 
D+E=    0;      "      E=+2. 
&c.,  &c. 

Substituting  these  values  of  the  coefficients  in  the  assumed 
series,  we  obtain 


<-,  &c. 
1+x 

(302.)  The  method  thus  exemplified  is  expressed  in  the  fol- 
lowing 

RULE. 

Assume  a  series  with  unknown  coefficients  as  equal  to  the  pro- 
posed expression  ;  then,  having  cleared  the  equation  of  fractions, 
or  raised  it  to  its  proper  power,  find  the  value  of  each  of  these 
coefficients  by  equating  the  corresponding  terms  of  the  two  ex- 
pressions,  or  putting  such  of  them  as  have  no  corresponding 
terms,  equal  to  zero. 

Ex.  2.  Expand  the  fraction  -  —  -  —  ;  —  r  into  an  infinite  series. 

1—  2x+x* 

Assume        0X       9=A+Ex+Cx*+Dx*+ftx'+,  &c. 
1  ~~~  £x~\~x 

Multiplying  by  1—  2x+x*,  we  have 


INFINITE    SERIES. 


C)x4+,  &c. 
Hence  we  must  have 

A=l 

B-2A=0  v  B=2A        =2, 
C-2B+  A=OvC=2B-A=3, 
D-2C+  B=0  v  D=2C-B=4, 
E-2D+  C=OvE=2D-C=5? 
&c.,  &c. 

Therefore,  —  ^-7-^=  l+2x+3x*  +4x3+5x4+  ,  &c. 


1    I   %x 

Ex.  3.  Expand  the  fraction  -  --  ;  into  an  infinite  series. 

1  -~ic     x 


,  &c., 

where  the  coefficient  of  each  term  is  equal  to  the  sum  of  the 
coefficients  of  the  two  preceding  terms. 

1       _    ,£ 

Ex.  4.  Expand  -  —  -  -  —  ^  into  an  infinite  series. 
1  —  <&x  —  ox 

Ans.  l+^4-5^2+13a;3+4l2:4+121x6+,  &c 
What  is  the  law  of  the  coefficients  in  this  series  ? 

1      I     O-y 

Ex.  5.  Expand  —  -    -  into  an  infinite  series. 
1  —  ox 


Ans.  l  +  5x  +  15xa+45x3+135o;4+,  &c. 
What  is  the  law  of  the  coefficients  in  this  series  1 


Ex.  6.  Expand  Vl—x  into  an  infinite  series. 

x     x*       3^3        3.5z4         3.5.7xs 
~2~2A~2A^~2A&8~2A.6.8.1Q~'      C' 

(303.)  The  method  of  unknown  coefficients  requires  that  we 
should  know  beforehand  the  form  of  the  development  with  re- 
spect to  the  powers  of  x.  Generally,  we  suppose  the  develop- 
ment to  proceed  according  to  the  ascending  powers  of  x,  com- 
mencing with  x° ;  but  sometimes  this  form  is  inapplicable,  in 
which  case  the  result  of  the  operation  is  sure  to  indicate  it. 

Let  it  be  required,  for  example,  to  develop  the  expression 

into  a  series. 

3x— x2 

Assume       — -—=A+Bx+Cx*+T>x>+,  &c. 


254  INFINITE    SERIES. 

Clearing  of  fractions,  we  have 


,  &c.  ; 
whence,  according  to  Art.  300,  we  conclude 

1=0, 

3A=0,  &c. 

Now  the  first  equation,  1=0,  is  absurd,  and  shows  that  the 
assumed  form  is  not  applicable  in  the  present  case.     But  if  we 

put  the  fraction  under  the  form  —  X«  -  ,  and  suppose  that 

x    3—  x 

±X^-=-(A.+Kx+Cx>+Vx*+,  &c.), 

X       O  —  X       X 

it  will  become,  after  the  reductions  are  made, 

l=3A+(3B-A)z+(3C-BX+(3D-C)z3-f,  &c., 
which  gives  the  equations 

3A=1  ;  whence  A=  J. 
3B-A=0;  "  B=J. 
3C-B=0;  "  C=sV 
3D-C=0;  "  D=TV 

Therefore,  §_l-J=i(I+|+g+g+,  &c.) 

x-1    x°     x   ,  x'  , 
=T+9-+27+81+'&C-; 

that  is,  the  development  contains  a  term  affected  with  a  nega- 
tive exponent. 

We  ought,  then,  to  have  assumed  at  the  outset 

,  &c. 


The  particular  series  which  should  be  adopted  in  each  case 
may  be  determined  by  putting  x=Q,  and  observing  the  nature 
of  the  result.  If,  in  this  case,  the  proposed  expression  becomes 
equal  to  a  finite  quantity,  the  first  term  of  the  series  will  not 

contain  x.     If  the  expression  reduces  to  zero,  the  first  term 

^ 
will  contain  x  ;  and  if  the  expression  reduces  to  the  form  •-, 

then  the  first  term  of  the  development  must  contain  x  with  a 
negative  exponent. 


SECTION  XIX. 


GENERAL  THEORY  OF  EQUATIONS. 

(304.)  It  is  proposed  in  this  Section  to  exhibit  the  most  im- 
portant propositions  relating  to  the  theory  of  equations,  to- 
gether with  the  Theorem  of  Sturm,  by  which  we  are  enabled 
to  determine  the  number  of  real  roots  of  an  equation. 

A  function  of  a  quantity  is  any  expression  involving  that 
quantity.  Thus, 

ax*+b  is  a  function  of  x. 
ay*+cy+d  is  a  function  of  y. 
ax*—  by*  is  a  function  of  a;  and  y. 

In  a  series  of  terms,  two  successive  signs  constitute  a  per- 
manence when  the  signs  are  alike,  and  a  variation  when  they 
are  unlike.     Thus,  in  the  polynomial 
a+b—  c+d, 

the  signs  of  the  first  two  terms  constitute  a  permanence  ;  the 
signs  of  the  second  and  third  constitute  a  variation  :  and  those 
of  the  third  and  fourth  also  a  variation. 

(305.)  A  cubic  equation  is  one  in  which  the  highest  power 
of  the  unknown  quantity  is  of  the  third  degree  ;  as,  for  ex- 
ample, 


All  equations  of  the  third  degree,  with  one  unknown  quan- 
tity, may  be  reduced  to  the  form 

x*+ax*+bx+c=0. 

A  biquadratic  equation  is  one  in  which  the  highest  power  of 
the  unknown  quantity  is  of  the  fourth  degree  ;  as,  for  example, 


12 


256  GENERAL  THEORY  OF  EQUATIONS. 

Every  equation  of  the  fourth  degree,  with  one  unknown 
quantity,  may  be  reduced  to  the  form 

x*+ax*+bx*+cx+d=Q. 
The  general  form  of  an  equation  of  the  fifth  degree  is 


and  the  general  form  of  an  equation  of  the  mth  degree,  with 
one  unknown  quantity,  is 

xm+Axm-1+Bxm-*+Cxm-3+  .....  +Tx+V=0  (m). 

This  equation  will  be  frequently  referred  to  hereafter  by  the 
name  of  the  general  equation  of  the  mth  degree,  or  simply  by  the 
letter  (m). 

It  is  obvious,  that  if  we  could  solve  this  equation,  we  should 
have  the  solution  of  every  equation  which  could  be  proposed. 
Unfortunately,  no  general  solution  has  ever  been  discovered  ; 
yet  many  important  properties  are  known,  which  enable  us  to 
solve  any  numerical  equation  which  can  ever  occur. 

PROPOSITION  I. 

(306.)  If  a  is  a  root  of  the  general  equation  of  the  mth  degree, 
the  equation  will  be  exactly  divisible  by  x—  a. 

For  if  a  is  one  value  of  x,  the  equation  must  be  verified  when 
we  substitute  a  in  the  place  of  x.     Hence  we  must  have 
am+Aam-1+Bam-2+Cam-3+  .....  +T«+V=0  (1). 

Subtracting  equation  (1)  from  equation  (m),  we  obtain 
(xm-am)+A(xm-l-am-l)+B(xm-*-am-*)  +  .  .  .  +  T(x-a)=0  (2). 

But,  by  Art.  76,  each  of  the  expressions  (xm—  am),  (xm~1—am~l), 
&c.,  is  divisible  by  x—a,  and  therefore  equation  (2)  is  also  di- 
visible by  x—a.  Now  equation  (m)  is  but  another  form  for 
equation  (2)  ;  for  if  we  take  the  value  of  V,  as  found  from 
equation  (1),  and  substitute  it  for  V  in  equation  (m),  it  will  give 
us  equation  (2)  ;  therefore,  equation  (m)  is  divisible  by  x—  a. 

Conversely,  if  equation  (m)  is  divisible  by  x—a,  then  a  is  a 
root  of  the  equation. 

It  will  be  noticed  that  this  property  is  but  a  generalization 
of  what  has  been  proved  of  equations  of  the  second  degree,  in 
Art.  192. 

Ex.  1.  Prove  that  1  is  a  root  of  the  equation 


GENERAL  THEORY  OF  EQUATIONS.  257 

This  equation  is  divisible  by  x—  1,  and  gives  #*—  5#+6=0. 
Ex.  2.  Prove  that  2  is  a  root  of  the  equation 
x*-x-6=0. 

This  equation  is  divisible  by  x—  2,  and  gives  x*+2x+3=Q. 
Ex.  3.  Prove  that  2  is  a  root  of  the  equation 


Ex.  4.  Prove  that  4  is  a  root  of  the  equation 

x*+x*—34x+56=0. 
Ex.  5.  Prove  that  —  1  is  a  root  of  the  equation 


Ex.  6.  Prove  that  —5  is  a  root  of  the  equation 

x5+6^4-  Wx*-  1  12x*-207x-  110=0. 
Ex.  7.  Prove  that  3  is  a  root  of  the  equation 


PROPOSITION  II. 

(307.)  Every  equation  of  the  mth  degree  containing  but  one 
unknown  quantity,  has  m  roots  and  no  more. 

For,  suppose  a  to  be  a  root  of  the  general  equation  of  the 
mth  degree.  By  the  last  Proposition,  this  equation  is  divisible 
by  x—a;  and  if  we  actually  perform  the  division,  the  equation 
will  be  reduced  to  one  of  the  next  inferior  degree. 

If  we  represent  the  coefficients  of  the  different  powers  of  x 
by  A',  B',  &c.,  the  quotient  will  be 


This  equation  must  also  have  a  root,  which  we  will  repre- 
sent by  b  ;  and  dividing  by  x—  fc,  the  equation  will  be  reduced 
to  one  of  the  next  inferior  degree,  and  so  on. 

We  may  continue  this  series  of  operations  (m—l)  times,  when 
we  shall  arrive  at  a  simple  equation  which  has  only  one  root. 
Hence  the  proposed  equation  will  have  m  roots, 

a,b,c,d,  .....  /; 

and  its  successive  divisors,  or  the  factors  of  which  it  is  com- 
posed, will  be 

x—a,  x—bt  x  —  c,  x—df  .....  x—  /, 

being  equal  in  number  to  the  units  contained  in  m,  the  highest 
exponent  of  the  equation. 

R 


258  GENERAL  THEORY  OF  EQUATIONS. 

We  have  seen  that  when  one  root  of  an  equation  is  known, 
the  equation  is  readily  reduced  to  one  of  the  next  inferior  de- 
gree ;  and  if  we  can  depress  any  equation  to  a  quadratic,  its 
roots  can  be  determined  by  methods  already  explained. 

Ex.  1.  One  root  of  the  equation 


is  1.     Find  the  remaining  roots. 
Ex.  2.  Two  roots  of  the  equation 

a:4-  10z3-f35:e'-502;+24=0 
are  1  and  3.     Find  the  remaining  roots. 
Ex.  3.  Two  roots  of  the  equation 

x*—  12a;3+48:ca-68:c-j-15=0 
are  3  and  5.     Find  the  remaining  roots. 

Ans. 
Ex.  4.  Two  roots  of  the  equation 


are  2  and  3.     Find  the  remaining  roots. 

Ans. 
Ex.  5.  Two  roots  of  the  equation 


are  2  and  —2.     Find  the  remaining  roots. 

Ans. 

(308.)  It  should  be  observed  that  this  Proposition  only  proves 
that  an  equation  of  the  mth  degree  may  be  continually  depress- 
ed by  division,  and  finally  exhausted  after  m  operations.     The 
divisors  are  not  necessarily  unequal.     Any  number,  and  indeed 
all  of  them,  may  be  equal.     When  we  say  that  an  equation  of 
the  mth  degree  has  m  roots,  we  mean  that  the  polynomial  can 
be  decomposed  into  m  binomial  factors,  equal  or  unequal,  each 
containing  one  root.     Thus,  the  equation 
x3-6a;a+12a;-8=0 
can  be  resolved  into  the  factors 

(x-2)  (x-2)  (x-2)=0;  or  (x-2)s=0; 
whence  it  appears  that  the  three  roots  of  this  equation  are 

2,  2,  2. 

But,  in  general,  the  several  roots  of  an  equation  differ  from 
each  other  numerically. 


GENERAL    THEORY    OF    EQUATIONS.  259 

The  equation 

x*=8 

has  apparently  but  one  root,  viz.,  2;  but  by  the  method  of  the 
preceding  article  we  can  discover  two  other  roots.  Dividing 
a;8—  8  by  x—2,  we  obtain 


Solving  this  equation,  we  find 


Thus,  the  three  roots  of  the  equation  x*=8  are 

2;  -l+V^S;  -l-V^3. 

These  last  two  values  may  be  verified  by  multiplication  as 
follows  : 

-1+  V^3  -I-    V]£3 

-1+  V-3  -1-    V-3 

1-  V~^3  1  +    V~-3 

-  V^-3-3  +    V  —  3—3 


—2—2  V  —  3=the  square.        —2+2  ^^-3= the  square. 
-1+    V^3  -1-    V^S 


3  2-2V-3 

-2V-3+6  +2^-3+6 


8—  the  cube.  8=  the  cube. 

If  the  last  term  of  an  equation  vanishes,  as  in  the  example 


the  equation  is  divisible  by  #—0,  and,  consequently,  0  is  one 
of  its  roots. 

If  the  last  two  terms  vanish,  then  two  of  its  roots  are  equal 
toO. 

PROPOSITION  III. 

To  discover  the  law  of  the  coefficients  of  every  equation. 
(309.)  In  order  to  discover  the  law  of  the  coefficients,  let  us 
form  the  equation  whose  roots  are 

«,  &,  c,  d,  .....  /. 

This  equation  will  contain  the  factors  (x—  a),  (x—  6),  (x—  c)t 
&c.  ;  that  is,  we  shall  have 

'•<:-«)  (x-b)  (x-c)  (x-d)  .....  (z-/)  =  0. 


260 


GENERAL  THEORY  OF  EdUATIONS. 


xm—  a 

x^+ab 

xm~*—  abc 

-b 

+  ac 

-abd 

—  c 

+ad 

—acd 

-d 

+  bc 

-bed 

&c.        +bd 

&c. 

+  cd 

If  we  perform  the  multiplication  as  in  Art.  264,  we  shall 
have 

+  .....  -(abc  .....  /)=0. 


i>  ow  ,§— -&  yti  a- -  •• 


&c. 
Hence  we  perceive, 

1.  The  coefficient  of  the  second  term  of  any  equation  is  equal 
to  the  sum  of  all  the  roots  with  their  signs  changed. 

2.  The  coefficient  of  the  third  term  is  equal  to  the  sum  of  the 
products  of  all  the  roots  taken  two  and  two. 

3.  The  coefficient  of  the  fourth  term  is  equal  to  the  sum  of  the 
products  of  all  the  roots  taken  three  and  three,  with  their  signs 
changed. 

4.  The  last  term  is  the  product  of  all  the  roots  with  their  signs 
changed. 

It  will  be  perceived  that  these  properties  include  those  of 
quadratic  equations  mentioned  on  pages  163  and  164. 

If  the  roots  are  all  negative,  the  signs  of  all  the  terms  of  the 
equation  will  be  positive,  because  the  factors  of  which  the 
equation  is  composed  are  all  positive. 

If  the  roots  are  all  positive,  the  signs  of  the  terms  will  be  al- 
ternately +  and  -. 

Ex.  1.  Form  the  equation  whose  roots  are  1,  2,  and  3. 

For  this  purpose,  we  must  multiply  together  the  factors 
a;—  1,  a;—  2,  x—3,  and  we  obtain 


This  example  conforms  to  the  rules  above  given  for  the  co- 
efficients. Thus,  the  coefficient  of  the  second  term  is  equal  to 
the  sum  of  all  the  roots  (1+2+3)  with  their  signs  changed. 

The  coefficient  of  the  third  term  is  the  sum  of  the  products 
of  the  roots  taken  two  and  two  ;  thus, 

1X2+1X3+2X3=11. 

The  last  term  is  the  product  of  all  the  roots  (1X2X3)  with 
their  signs  changed. 


GENERAL  THEORY  OF  EQUATIONS.  261 

Ex.  2.  Form  the  equation  whose  roots  are  2,  3,  5,  and  —6. 

Ans.  x*  —4x*-29x*+  156^-1  80=0. 

Show  how  these  coefficients  conform  to  the  laws  above 
given. 

Ex.  3.  Form  the  equation  whose  roots  are  1,  3,  5,  —2,  —4, 
-6. 

Ans.  x6+3x6-41x*-87x*+4QOx*-4443;-1l20=0. 
(310.)  Every  rational  root  of  an  equation  is  a  divisor  of  the 
last  term  ;  for,  since  this  term  is  the  product  of  all  the  roots,  it 
must  be  divisible  by  each  of  them.     If,  then,  we  wish  to  find  a 
root  by  trial,  we  know  at  once  what  numbers  we  must  employ. 
For  example,  take  the  equation 

X*-X-G=0. 

If  this  equation  has  a  rational  root,  it  must  be  a  divisor  of 
the  last  term,  6  ;  hence  we  must  try  the  numbers  1,  2,  3,  6, 
either  positive  or  negative. 

If  x=I,  we  have       1  —  1—6=  —  6, 

x=2,        "  8-2-6=     0, 

z=3,        "  27-3-6=   18, 

#=6,        "  216-6-6=204, 

Hence  we  see  that  2  is  one  of  the  roots  of  the  given  equa- 
tion, and  by  the  method  of  Art.  307,  we  shall  find  the  remain- 
ing roots  to  be 


PROPOSITION  IV. 

(311.)  No  equation  whose  coefficients  are  all  integers,  and  that 
of  the  first  term  unity,  can  have  a  root  equal  to  a  rational  frac- 
tion. 

For,  take  the  general  equation  of  the  third  degree, 


and  suppose,  if  possible,  that  the  fraction  r  is  one  value  of  x, 

this  fraction  being  reduced  to  its  lowest  terms.     If  we  substi- 
tute this  value  for  x  in  the  given  equation,  we  shall  have 


262  GENERAL  THEORY  OF  EaUATIONS. 

Multiplying  each  term  by  6a,  and  transposing,  we  obtain 


Now,  by  supposition,  A,  B,  C,  a  and  b  are  whole  numbers. 
Hence  the  entire  right-hand  member  of  the  equation  is  a  whole 
number. 

But  by  hypothesis,       is  an  irreducible  fraction  ;  that  is,  a 


and  b  contain  no  common  factor.     Consequently,  a9  and  b  will 


contain  no  common  factor,  that  is,  T  is  a  fraction  in  its  lowest 


terms.     Hence,  if  7-  were  a  root  of  the  proposed  equation,  we 

should  have  a  fraction  in  its  lowest  terms  equal  to  a  whole 
number,  which  is  absurd. 

The  same  mode  of  demonstration  is  applicable  to  the  general 
equation  of  the  rath  degree. 

This  proposition  only  asserts  that  in  an  equation  such  as  is 
here  described,  the  real  roots  must  be  integers,  or  they  can  not 
be  exactly  expressed  in  numbers.  They  may  often  be  express- 
ed approximately  by  fractions,  as  is  seen  in  the  examples  on 
pages  288-301.  A  real  root  which  can  not  be  exactly  ex- 
pressed in  numbers  is  called  incommensurable. 

PROPOSITION  V. 

(312.)  If  the  signs  of  the  alternate  terms  in  an  equation  are 
changed,  the  signs  of  all  the  roots  will  be  changed. 

If  we  take  the  general  equation  of  the  mth  degree,  and  change 
the  signs  of  the  alternate  terms,  we  shall  have 

xm-Axm-1+Rxm-'-Cxn-*+  .....  =0     (1)  : 
or,  changing  the  sign  of  every  term  of  the  last  equation, 
-x^+Ax^-Bx^+Cx"-*-  .....  =0     (2). 

Now,  substituting  +a  for  x  in  equation  (m)  will  give  the 
same  result  as  substituting  —a  in  equation  (1),  if  m  be  an  even 
number;  or,  substituting  —am  equation  (2),  if  m  be  an  odd 
number.  If,  then,  a  is  a  root  of  equation  (m),  —a  will  be  a  root 
of  equation  (1),  and,  of  course,  a  root  of  equation  (2),  which  is 
identical  with  it. 


GENERAL  THEORY  OF  EQUATIONS.  263 

Hence  we  see  that  the  positive  roots  may  be  changed  into 
negative  roots,  and  the  reverse,  by  simply  changing  the  signs 
of  the  alternate  terms  ;  so  that  the  finding  the  real  roots  of  any 
equation  is  reduced  to  finding  positive  roots  only. 

Ex.  1.  The  roots  of  the  equation 


are  1,  3,  and  —  2.     What  are  the  roots  of  the  equation 

x*+2x*-5x-6=0? 
Ex.  2.  The  roots  of  the  equation 

x*-6x*+llx-6=0 
are  1,  2,  and  3.     What  are  the  roots  of  the  equation 


PROPOSITION  VI. 

(313.)  If  an  equation  whose  coefficients  are  all  real,  contains 
imaginary  roots,  the  number  of  these  roots  must  be  even. 

If  an  equation  whose  coefficients  are  all  real,  has  a  root  of 
the  form 


then  will  a—  6V  —  1 

be  also  a  root  of  the  equation. 

For,  let  a-\-bV  —  1  be  substituted  for  x  in  the  equation,  the 
result  will  consist  of  a  series  of  terms,  of  which  those  involving 

only  the  powers  of  a,  and  the  even  powers  of  bV  —  1  will  be 
real,  and  those  which  involve  the  odd  powers  of  bV  —  1  will 
be  imaginary.  If  we  denote  the  sum  of  the  real  terms  by  P, 

and  the  sum  of  the  imaginary  terms  by  Qv7  —  1,  then  we  must 
have 


which  relation  can  only  exist  when  P=0  and  Q=0. 

Again,  let  a—bV  —  l  be  substituted  for  x  in  the  proposed 
equation,  the  only  difference  in  the  result  will  be  in  the  signs 
of  the  odd  powers  of  bV  —  1,  so  that  the  result  will  be  P— 
Q,  V  —  1.  But  we  have  found  that  P=0  and  Q=0  ;  hence 


12 


264  GENERAL  THEORY  OF  EQUATIONS. 

And,  since  a—  b  V  —  1  substituted  for  x  gives  a  result  equal  to 
zero,  it  must  be  a  root  of  the  equation. 
Ex.  1.  Find  the  roots  of  the  equation 


Ans.  —2,  and  1±-/  —  1. 

Ex.  2.  Find  the  roots  of  the  equation 


Ans.  —3,  and  2±  V  —  1. 
Hfc.  3.  Find  the  roots  of  the  equation 
5x*+2x—  44=0. 

'A-T      -•'        M  ~  1,1  /  -  «  - 

.Arcs.  2,  and  -1±V—  3.4. 

Hence  every  equation  of  the  third  degree  whose  coefficients 
are  all  real,  must  have  one  real  root.  The  same  is  true  of 
every  equation  of  an  odd  degree. 

•  i  U«  sm>  i^$H!ft*o'i  oio&B  i'.omws  £!£}• 

PROPOSITION  VII. 

(314.)  Every  equation  must  have  as  many  variations  of  sign 
as  it  has  positive  roots,  and  as  many  permanences  of  sign  as 
there  are  negative  roots. 

To  prove  this  Proposition,  it  is  only  necessary  to  show  that 
the  multiplication  of  an  equation  by  a  new  factor,  x—  a,  cor- 
responding to  a  positive  root,  will  introduce  at  least  one  varia- 
tiont  and  that  the  multiplication  by  a  factor  x+a  will  intro- 
duce at  least  one  permanence. 

T-*  ,  , 

For  an  example,  take  the  equation 


in  which  the  signs  are  +H ,  giving  one  variation. 

Multiply  this  equation  by  a;— 2=0,  as  follows  : 

x*+3x*-10x  -24 
x-2 

il38t!<|OU[    3i. 

z*+  x'-16x*-  4^+48=0. 

In  this  last  product  the  signs  are  +H h,  giving  two  va- 
riations ;  that  is,  the  introduction  of  a  positive  root  has  intro- 
duced one  new  variation  in  the  signs  of  the  terms. 


GENERAL  THEORY  OF  EaUATIONS.  265 

To  generalize  this  reasoning,  we  perceive  that  the  signs  in 
the  upper  line  of  the  partial  products  are  the  same  as  in  the 
given  equation  ;  but  those  in  the  lower  line  are  all  contrary  to 
those  of  the  given  equation,  and  advanced  one  term  toward  the 
right. 

Now,  if  each  coefficient  of  the  upper  line  is  greater  than  the 
corresponding  one  in  the  lower,  the  signs  of  the  upper  line  will 
be  the  same  as  in  the  total  product,  with  the  exception  of  the 
last  term.  But  the  last  term  introduces  a  new  variation,  since 
its  sign  is  contrary  to  that  which  immediately  precedes  it; 
that  is,  the  product  contains  one  more  variation  than  the 
original  equation. 

When  a  term  in  the  lower  line  is  larger  than  the  correspond- 
ing one  in  the  upper  line,  and  has  the  contrary  sign,  there  is  a 
change  from  a  permanence  to  a  variation  ;  for  the  lower  sign 
is  always  contrary  to  the  preceding  upper  sign.  Hence,  when- 
ever we  are  obliged  to  descend  from  the  upper  to  the  lower 
line  in  order  to  determine  the  sign  of  the  product,  there  is  a 
variation  which  is  not  found  in  the  proposed  equation  ;  and  as 
all  the  remaining  signs  of  the  lower  line  are  contrary  to  those 
of  the  proposed  equation,  there  must  be  the  same  changes  of 
sign  in  this  line  as  in  the  given  equation.  If  we  are  obliged  to 
reascend  to  the  upper  line,  the  result  may  be  either  a  variation 
or  a  permanence.  But  even  if  it  were  a  permanence,  since 
the  last  sign  of  the  product  is  in  the  lower  line,  it  is  necessary 
to  go  once  more  from  the  upper  line  to  the  lower,  than  from 
the  lower  to  the  upper.  Hence  each  factor,  corresponding  to 
a  positive  root,  must  introduce  at  least  one  new  variation  ;  so 
that  there  must  be  as  many  variations  as  there  are  positive 
roots. 

In  the  same  manner,  we  may  prove  that  the  multiplication 
by  a  factor  x-\-a,  corresponding  to  a  negative  root,  must  intro- 
duce at  least  one  new  permanence  ;  so  that  there  must  be  as 
many  permanences  as  there  are  negative  roots. 

Ex.  1.  The  roots  of  the  equation 


are  1,  2,  3,  —1,  and  —2.  There  are  also  three  variations  of 
sign,  and  two  permanences,  as  there  should  be,  according  to 
the  Proposition. 


266  GENERAL  THEORY  OF  EQUATIONS. 

Ex.  2.  The  equation 


has  four  real  roots.     How  many  of  these  are  negative  ? 
Ex.  3.  The  equation 


has  six  real  roots.     How  many  of  these  are  positive  ? 

If  all  the  roots  of  an  equation  are  real,  the  number  of  posi- 
tive roots  must  be  the  same  as  the  number  of  variations,  and 
the  number  of  negative  roots  must  be  the  same  as  the  number 
of  permanences.  If  any  term  of  an  equation  is  wanting,  we 
must  supply  its  place  with  ±0  before  applying  the  preceding 
Rule. 

PROPOSITION  VIII. 

(315.)  If  two  numbers,  when  substituted  for  the  unknown 
quantity  in  an  equation,  give  results  with,  contrary  signs,  there 
is  at  least  one  root  comprised  between  those  numbers. 

Take,  for  example,  the  equation 

x*—2x*+3x—  44=0. 

If  we  substitute  3  for  x  in  this  equation,  we  obtain  —26  ; 
and  if  we  substitute  5  for  x,  we  obtain  +46.  There  must, 
therefore,  be  a  real  root  between  3  and  5  ;  for,  when  we  sup- 
pose x=3,  we  have 


But  when  we  suppose  x=5,  we  have 


Now  both  the  quantities 

x*+3x  and  2#a+44 

increase  while  x  increases.  And  since  the  first  of  these  quan- 
tities, which  was  originally  less  than  the  second,  has  become 
the  greater,  it  must  increase  more  rapidly  than  the  second. 
There  must,  therefore,  be  a  point  at  which  the  two  magnitudes 
are  equal,  and  that  value  of  x  which  renders  these  two  magni- 
tudes equal  must  be  a  root  of  the  proposed  equation. 

In  general,  if  two  numbers,  p  and  q,  substituted  for  x  in  an 
equation,  give  results  with  contrary  signs,  we  may  suppose  the 
less  of  the  two  numbers  to  increase  by  imperceptible  degrees 


GENERAL  THEORY  OF  EQUATIONS.  26"? 

until  it  becomes  equal  to  the  greater  number.  The  results  of 
these  successive  substitutions  must  also  change  by  impercepti- 
ble degrees,  and  must  pass  through  all  the  intermediate  values 
between  the  two  extremes.  But  the  two  extreme  values  are 
affected  with  opposite  signs  ;  there  must,  therefore,  be  some 
number  between  p  and  q  which  reduces  the  given  equation  to 
zero,  and  this  number  will  be  a  root  of  the  equation. 

In  the  same  manner,  it  may  be  proved  that  if  any  quantity 
p,  and  every  quantity  greater  than  p,  substituted  in  an  equation, 
renders  the  result  positive,  then  p  is  greater  than  the  greatest 
root. 

Hence,  also,  if  the  signs  of  the  alternate  terms  are  changed, 
and  if  q,  and  every  quantity  greater  than  q,  renders  the  result 
positive,  then  —  q  is  less  than  the  least  root. 

If  the  two  numbers,  which  give  results  with  contrary  signs, 
differ  from  each  other  only  by  unity,  it  is  plain  that  we  have 
found  the  integral  part  of  a  root. 

Ex.  1.  Find  the  integral  part  of  one  of  the  roots  of  the  equa- 
tion 


When  x  =2,  the  equation  reduces  to  —12;  and  when  x  =3, 
it  reduces  to  +71.  Hence  there  must  be  a  root  between  2  and 
3  ;  that  is,  2  is  the  first  figure  of  one  of  the  roots. 

Ex.  2.  Find  the  first  figure  of  one  of  the  roots  of  the  equa- 
tion 

x*+x*+x-  100=0. 

Ans.  4. 

Ex.  3.  Find  the  first  figure  of  each  of  the  roots  of  the  equa- 
tion 


PROPOSITION  IX. 

(316.)  Every  equation  may  be  transformed  into  another, 
whose  roots  are  greater  or  less  than  those  of  the  former  by  any 
given  quantity. 

Let  it  be  required  to  transform  the  general  equation  of  the 
mth  degree  into  another  whose  roots  are  greater  by  r  than 
those  of  the  given  equation. 

Take  y=x+r,  or  x=y—rt 


268 


GENERAL    THEORY    OF    EQUATIONS. 


and  substitute  y— r  for  x  in  the  proposed  equation;  we  shall 
then  have 

i-l)(w-2)r8 


+A 


-(w-l)Ar 
+B 


2T-2- 


2.3 

(m-l)(m-2)Ar* 


-(m-2)Br 
+C 

which  equation  evidently  fulfills  the  required  conditions,  smce 
y  is  greater  than  x  by  r. 

If  we  take  y=x—r,  or  x—y-\-r,  we  shall  obtain  in  the  same 
way  an  equation  whose  roots  are  less  than  those  of  the  given 
equation  by  r. 

Ex.  1.  Find  the  equation  whose  roots  are  greater  by  1  than 
those  of  the  equation 


We  must  here  substitute  y—  I  in  place  of  x. 

Ans.  y- 

Ex.  2.  Find  the  equation  whose  roots  are  less  by  1  than 
those  of  the  equation 

x*-2x*+Zx-4=0. 

Ans.  y-fy9+2y-2=0. 

Ex.  3.  Find  the  equation  whose  roots  are  greater  by  3  than 
those  of  the  equation 

x'+9x*+I2x*-14x=0. 

Ans.  y4—  3?/8—  15y3+49y—  12=0. 

Ex.  4.  Find  the  equation  whose  roots  are  less  by  2  than 
those  of  the  equation 


Ans.  5 

Ex.  5.  Find  the  equation  whose  roots  are  greater  by  2  than 
those  of  the  equation 


Ans. 


PROPOSITION  X. 

(317.)  Any  complete  equation  may  be  transformed  into  an- 
other whose  second  term  is  wanting. 


GENERAL  THEORY  OF  EQUATIONS.  269 

Since  r  in  the  preceding  Proposition  is  indeterminate,  we 

may  put  —  mr+A  equal  to  zero,  which  will  cause  the  second 

^ 
term  of  the  general  development  to  disappear.     Hence  r=—  , 

A 

and  x=y  --  . 
y     m 

Hence,  to  remove  the  second  term  of  an  equation,  substitute 
for  the  unknown  quantity  a  new  unknown  quantity,  together 
with  such  a  part  of  the  coefficient  of  the  second  term,  taken  with 
a  contrary  sign,  as  is  denoted  by  the  degree  of  the  equation. 

Ex.  1.  Transform  the  equation 


into  another  whose  second  term  is  wanting. 

Here  we  take  a  new  unknown  quantity,  and  annex  to  it  a 
third  part  of  the  coefficient  of  the  second  term  of  the  equation 
with  its  sign  changed  ;  that  is,  we  put  x=y+2.  Making  this 
substitution,  we  obtain 

ys—4y—2=0.  Ans. 

Ex.  2.  Transform  the  equation 


into  another  whose  second  term  is  wanting. 
Here  we  put  x=y+4. 

Ans.  y*—96y*—518y—  777=0. 
Ex.  3.  Transform  the  equation 


into  another  whose  second  term  is  wanting. 
Ans.     5- 


Since  the  coefficient  of  the  second  term  is  equal  to  the  sum 
of  the  roots  with  their  signs  changed,  it  is  obvious  that  when 
the  second  term  of  an  equation  is  wanting,  the  sum  of  the  posi- 
tive roots  must  be  equal  to  the  sum  of  the  negative  roots. 

PROPOSITION  XL 

(318.)   To  discover  the  law  of  Derived  Polynomials. 
When  we  substitute  y+r  for  x  in  the  general  equation  of 
the  with  degree,  the  coefficients  of  r  follow  a  remarkable  law. 
The  equation,  before  it  is  developed,  is 


270  GENERAL  THEORY  OF  EQUATIONS. 


'n-a+  .....  +T(y+r)+V=0. 
If  we  actually  involve  the  several  terms  (y+r)m,  (y+r)'*"1, 
&c.,  as  was  done  in  Art.  316,  we  obtain  certain  terms  inde- 
pendent of  r,  others  which  contain  the  first  power  of  r,  others 
the  second  power  of  r,  and  so  on  ;  and  the  development  is  of 
the  following  form  : 


where  the  values  of  X,  Xlf  X2,  &c.,  are 
X  =ym+Aym-l+Bym-*+Cym-*+  .....  +  Ty+V. 


Each  of  these  polynomials  may  be  derived  from  that  imme- 
diately preceding  it,  by  multiplying  each  term  by  the  exponent 
of  y  in  that  term,  and  diminishing  the  exponent  by  unity. 

The  expressions  Xlf  X2,  &c.,  are  called  derived  polyno- 
mials of  X.  Xj  is  called  the  first  derived  polynomial,  X2  the 
second  derived  polynomial,  X3  the  third,  and  so  on. 

Ex.  1.  Find  the  equation  whose  roots  are  less  by  r  than 
those  of  the  equation 


Here  we  shall  have 

X  =  ya-5y+6, 


But  we  have  seen  that  when  y+r  is  substituted  for  x,  the 
equation  reduces  to  the  form 


Substituting  the  values  of  X,  X,,  X2,  &c.,  above  found,  we 
obtain 


which  is  the  development  of 


Ex.  2.  Find  the  equation  whose  roots  are  less  by  /•  than 
those  of  the  equation 


GENERAL  THEORY  OF  EQUATIONS.  271 


Here  we  shall  have 

X  =  y8-  7ya+8y-3, 
X  ,=3^-1%  +8, 


X4=0; 

and,  substituting  these  values  in  the  same  formula  as  above, 
we  obtain 


which  is  the  development  of 


Ex.  3.  Find  the  successive  derived  polynomials  of  the  equa- 
tion 


Ex.  4.  Find  the  successive  derived  polynomials  of  the  equa- 
tion 

x6+3x4+2xs—3x*-2x—2=0. 

PROPOSITION  XII. 

(319.)   To  find  the  equal  roots  of  an  equation. 
We  have  seen,  in  Art.  308,  that  an  equation  may  have  two 
or  more  equal  roots.     Thus,  the  equation 


or  (x-2)3=0, 

has  the  three  equal  roots  2,  2,  2.  Such  an  equation  and  its 
first  derived  polynomial  always  contain  a  common  divisor  ;  for 
the  first  derived  polynomial  of  the  above  equation  is 


or  3(#-2)a, 

where  it  is  evident  that  (x—  2)a  is  a  common  divisor  of  both 

equations. 

In  general,  let  a  be  one  of  the  equal  roots  which  occurs  n 
times  as  a  root  of  the  given  equation  ;  the  first  member  will 
therefore  contain  the  factors  (x—a),  (x—a),  (x—a),  —  -  ; 
that  is,  (x—  a)".  The  first  derived  polynomial  will  contain  the 
factor  n(x  —  a)"  '  ;  that  is,  x  —  a  occurs  (n—  1)  times  as  a  factor 


272  GENERAL  THEORY  OF  EQUATIONS. 

in  the  first  derived  polynomial.  The  greatest  common  divisor 
of  the  given  equation  and  its  first  derived  polynomial  must 
therefore  contain  the  factor  (x—  a)  repeated  once  less  than  in 
the  given  equation. 

To  determine,  therefore,  whether  an  equation  has  equal 
roots,  fin  d  the  greatest  common  divisor  between  the  equation  and 
its  first  derived  polynomial.  If  there  is  no  common  divisor,  the 
equation  has  no  equal  roots.  If  there  is  a  common  divisor,  solve 
the  equation  obtained  by  putting  this  divisor  equal  to  zero. 

Ex.  1.  Find  the  equal  roots  of  the  equation 


The  first  derived  polynomial  of  this  equation  is 


The  greatest  common  divisor  between  this  and  the  given 
equation  is 

x-3. 

Hence  the  equation  has  two  roots,  each  equal  to  3. 
Ex.  2.  Find  the  equal  roots  of  the  equation 


Ans.  Two  roots  equal  to  5. 
Ex.  3.  Find  the  equal  roots  of  the  equation 


Ans.  Two  roots  equal  to  2. 
Ex.  4.  Find  the  equal  roots  of  the  equation 
x'-6x*-8x-3=0. 

Ans.  Three  roots  equal  to—  1 

PROPOSITION  XIII. 

(320.)  To  find  the  number  of  real  and  imaginary  roots  of  an 
equation. 

In  1829,  M.  Sturm  discovered  a  theorem  which  determines 
the  precise  number  of  real  roots,  and  of  course  the  number  of 
imaginary  ones,  since  the  real  and  imaginary  roots  are  to- 
gether equal  in  number  to  the  degree  of  the  equation.  We 
propose  now  to  develop  this  theorem. 

Let  X  represent  the  first  member  of  the  general  equation  of 
the  mth  degree,  which  we  suppose  to  have  no  equal  roots,  and 


GENERAL  THEORY  OF  EQUATIONS.  273 

let  X,  be  its  first  derived  polynomial,  found  by  the  method  of 
Art.  318. 

Divide  X  by  X,  until  the  remainder  is  of  a  lower  degree 
than  the  divisor,  and  call  this  remainder  —  X77 ;  that  is,  let  X/7 
designate  the  remainder  with  a  contrary  sign.  Divide  X7  by 
X77  in  the  same  manner,  and  so  on,  designating  the  successive 
remainders  with  contrary  signs  by  X//7,  X/7//,  &c.,  until  the  di- 
vision terminates  by  leaving  a  numerical  remainder  independ- 
ent of  x ;  which  must  always  be  the  case,  according  to  the  pre- 
ceding Proposition,  since  the  equation  having  no  equal  roots, 
there  can  be  no  factor,  which  is  a  function  of  x,  common  to 
the  equation  and  its  first  derived  polynomial.  Let  this  re- 
mainder, having  its  signs  changed,  be  called  X». 

The  operation  thus  described  will  stand  as  follows : 


X      |X,  X, 

X,Q,Q,  X,, 


x//  x/7 


X 


/7/ 


Q,,, 


X— X7Q7— —  X77;   X,— X77Q/7— — X//7 ;   X/7— X7//Q,777 — — X7777. 

We  thus  obtain  the  series  of  quantities 

X,  X7,  X7/,  X/7/,  X///7, Xm, 

each  of  which  is  of  a  lower  degree  with  respect  to  x  than  the 
preceding,  and  the  last  is  altogether  independent  of  x,  that  is, 
does  not  contain  x. 

We  now  substitute  for  x  in  the  above  functions  any  two 
numbers  p  and  q,  of  which  p  is  less  than  q.  The  substitution 
of  p  will  give  results  either  positive  or  negative.  If  we  only 
take  account  of  the  signs  of  the  results,  we  shall  obtain  a  certain 
number  of  variations  and  a  certain  number  of  permanences. 

The  substitution  of  q  for  x  will  give  a  second  series  of  signs, 
presenting  a  certain  number  of  variations  and  permanences. 
The  following,  then,  is 

THE  THEOREM  OF  STURM. 

The  difference  between  the  number  of  variations  of  the  first 
row  of  signs  and  that  of  the  second,  is  equal  to  the  number  of 
real  roots  of  the  given  equation  comprised  between  p  and  q. 

(321.)  In  order  to  simplify  the  demonstration  of  this  theorem, 
we  shall  premise  three  Lemmas  ;  and,  for  convenience,  we  shall 
call  X  the  primitive  function,  and  X7,  X77,  X/7/,  &c.,  auxiliary 
functions. 

S 


GENERAL  THEORY  OF  EQUATIONS. 

LEMMA  I.  If  we  substitute  any  number  for  x  in  the  series  of 
functions  X,  X,,  X//5  &c.,  two  consecutive  functions  can  not  both 
reduce  to  zero  at  the  same  time. 

For,  from  the  method  in  which  X/,  X/;,  &c.,  are  obtained, 
we  have  the  following  equations  : 

X  =X,  Q,  -X,,  (1). 
X,  =X//Q/,  -X,,,  (2). 
X//  :=X///Q,///  —  X,,,,  (3). 


=Xm_1QOT_1—  XM       (w—  1). 

Now,  if  possible,  suppose  X/=0,  and  X//=0  ;  then,  by  equa- 
tion (2),  we  shall  have  X/,^0.  Also,  since  X/^0,  and  X7//= 
0  ;  therefore,  by  equation  (3),  we  must  have  X/,,^0  ;  and,  pro- 
ceeding in  the  same  manner,  we  shall  find  that  Xm=0,  which  is 
absurd,  since  it  was  shown,  Art.  320,  that  this  final  remainder 
must  be  independent  of  x,  and  must  therefore  remain  un- 
changed for  every  value  of  x. 

LEMMA  II.  'When  one  of  the  auxiliary  functions  vanishes  for 
a  particular  value  of  x,  the  two  adjacent  functions  must  have 
contrary  signs. 

For,  by  equation  (3),  we  have 

X//=A.///(qJ,///        -A-////  » 

and  if  X//7  reduces  to  zero,  then  X,,^—  X/7//  ;  that  is,  X/7  and 
X7///  have  contrary  signs. 

LEMMA  III.  If  &  is  a  root  of  the  equation  X=0,  the  signs  of 
X  and  X/  will  constitute  a  variation  for  a  value  of  x  which  is 
a  little  less  than  a,  and  a  permanence  for  a  value  ofx  which  is  a 
little  greater  than  a. 

For  if  we  substitute  a+r  for  x  in  the  equation  X=0,  the  de- 
velopment of  the  function  X,  according  to  Art.  318,  will  be  of 
the  form 

A+AV+  other  terms  involving  higher  powers  of  r. 

Now  if  a  is  a  root  of  the  equation  X=0,  the  first  term  of  the 
development  becomes  zero,  and  there  remains 

AV+  other  terms  involving  higher  powers  of  r. 

Also,  if  we  substitute  a+r  for  x  in  the  first  derived  polyno- 
mial, the  development  will  contain 

A'  4-  other  terms  involving  r. 


GENERAL  THEORY  OP  EQUATIONS.  275 

Now  we  may  take  r  so  small  that  each  of  these  develop- 
ments shall  have  the  same  sign  as  its  first  term, 

A'r  and  A'. 

Hence  they  must  both  have  the  same  sign  when  a  is  positive, 
and  contrary  signs  when  r  is  negative.     That  is,  the  signs  of 
the  two  functions  X  and  X, 
constitute  a  variation  for    x=a— r, 
and  a  permanence  for        x=a+r. 

DEMONSTRATION  OF  THE  THEOREM. 

(322.)  Suppose  all  the  real  roots  of  the  equations 

X=0,  X7=0,  X,,=:0,  X///=0,  &c., 

to  be  arranged  in  a  series  in  the  order  of  magnitude,  beginning 
with  the  least.  Let  p  be  less  than  the  least  of  these  roots,  and 
let  it  increase  continually  until  it  becomes  equal  to  q,  which 
we  suppose  to  be  greater  than  the  greatest  of  these  roots. 
Now  so  long  as  p  is  less  than  any  of  the  roots,  no  change  of 
signs  will  occur  from  the  substitution  of  p  for  x  in  any  of  these 
functions,  Art.  315  ;  but  when  p  arrives  at  a  root  of  any  of  the 
auxiliary  equations,  its  substitution  for  x  reduces  that  polyno- 
mial to  zero,  and  neither  the  preceding  nor  succeeding  func- 
tion can  vanish  for  the  same  value  of  a:  (Lemma  I.),  and  these 
two  adjacent  functions  have  contrary  signs  (Lemma  II.). 
Hence  the  entire  number  of  variations  of  sign  is  not  affected 
by  the  vanishing  of  any  of  the  auxiliary  functions ;  for  the 
three  adjacent  functions  must  reduce  to 
+,  0,  -,  or  -,  0,  +. 

Here  is  one  variation,  and  there  will  also  be  one  variation 
if  we  supply  the  place  of  the  0  with  either  +  or  —  ;  thus, 

+,  +,  — ,  or  — ,  +,  +, 
+,  -,  -,  or  — ,  -,  +. 

Suppose,  now,  p  to  pass  from  a  number  very  little  smaller, 
to  a  number  very  little  greater  than  a  root  of  the  primitive 
equation 

X=0, 

the  sign  of  X  will  be  changed  from  +  to  — ,  or  from  —  to  +, 
Art.  315.  The  signs  of  X  and  X/  constitute  a  variation  before 
the  change,  and  a  permanence  after  the  change  (Lemma  III.} 


27ft  GENERAL   THEORY    OF    EQUATIONS. 

Hence  the  change  of  sign  of  the  function  X  occasions  a  loss 
of  one  variation  of  sign. 

Again,  while  p  increases  from  a  number  very  little  smaller 
to  a  number  very  little  greater  than  another  root  of  the  equa- 
tion X=0,  a  second  variation  will  be  changed  into  a  perma- 
nence, and  so  on  for  the  other  roots  of  the  primitive  equation. 

Now,  since  all  the  real  roots  must  be  comprised  within  the 
limits  —oo  and  +  00  ,  if  we  substitute  these  values  for  x  in  the 
series  of  functions  X,  X/5  &c.,  the  number  of  variations  lost 
will  indicate  the  whole  number  of  real  roots.  A  third  suppo- 
sition, that  x=0,  will  show  how  many  of  these  roots  are  posi- 
tive and  how  many  negative  ;  and  if  we  wish  to  determine 
smaller  limits  of  the  roots,  we  must  try  other  numbers.  It  is 
generally  best  to  make  trial  in  the  first  instance  of  such  num- 
bers as  are  most  convenient  in  computation,  as,  1,  2,  10,  &c. 

EXAMPLES. 
(323.)  Ex.  1.  How  many  real  roots  has  the  equation 


Here  we  have         X.,=3x* 
Dividing  xa—  6x*+llx—  6  by  3x*—  12z-hll,  as  in  the  meth- 
od for  finding  the  greatest  common  divisor,  Art.  251,  we  have 
for  a  remainder  —  2#+4.     Hence,  rejecting  the  factor  2,  X// 
=x—  2.     Dividing  X,  by  X,,,  we  have  for  a  remainder  —  1. 
Therefore,  X/;/=  +  l. 
Hence  we  have 

X    =  xa-  6a:a+llz-6. 
X,  =3x*-l2x  +11. 

•A-//     ==        X     ~~         <£• 

x///=+i. 

If  we  substitute  —  GO  for  #  in  the  first  polynomial  x9—  6x*+ 
llx—  6,  the  sign  of  the  result  is  —  ;  substituting  —  oo  for  x  in 
the  second  polynomial  3x2—  12#+11,  the  sign  of  the  result  is 
-j-  ;  substituting  the  same  in  x—2,  the  sign  of  the  result  is 
—  ;  and  X//7,  being  independent  of  x,  will  remain  +  for  every 
value  of  x,  so  that  by  supposing  x=  —  oo  ,  we  obtain  the  series 
of  signs 

-  +  -  +. 

Proceeding  in  the  same  manner  for  other  assumed  values 
of  x,  we  shall  obtain  the  following  results  : 


GENERAL  THEORY  OF  EQUATIONS.  277 

Assumed  Values  of  x.  Resulting  Signs.  Variations. 

—  oo  —  -1 h  giving  3  variations. 

0  1 1-  «      3         « 

+  .9  h  •      +  "3         " 

+  1  0   +  -  +  "      2 

+  1.1  +  +  -  +  "2 

+  1.9  H h  "2         " 

+2  0   -   0   +  "       1         " 

+2.1  h  +  "1          " 

+2.9  -  +  +  +  "1 

+  3  0   +  +  +  "      0 

+  3.1  +  +  +  +  "0 

-{-          00  _|_        _j_        _|-        _{-  "0  " 

Here  the  three  roots  of  this  equation  are  seen  to  be  1,  2,  3, 
and  no  change  of  sign  in  either  function  occurs  by  the  substi- 
tution for  x  of  any  number  less  than  1  ;  but  when  p  exceeds  1, 
there  is  a  change  of  sign  in  the  original  equation  from  —  to 
+,  by  which  one  variation  is  lost.  When  p=2,  two  of  the 
functions  disappear  simultaneously,  showing  that  2  is  a  root  of 
the  second  derived  function  as  well  as  of  the  original  equation, 
and  a  second  variation  of  sign  is  lost.  Also,  when  p  becomes 
equal  to  3,  a  third  variation  is  lost ;  and  there  are  no  further 
changes  of  sign  arising  from  the  substitution  of  any  numbers 
between  3  and  +00  . 

There  are  three  changes  of  sign  of  the  primitive  function,  two 
of  the  first  auxiliary  function,  and  one  of  the  second  auxiliary 
function  ;  but  no  variation  is  lost  by  the  change  of  sign  of  any 
of  the  auxiliary  functions  ;  while  every  change  of  sign  of  the 
primitive  function  occasions  a  loss  of  one  variation. 

Ex.  2.  How  many  real  roots  has  the  equation 


Here  we  find 

X    =  x3- 

X,   =3x*-Wx  +8. 

X,,=2x-3l. 

X^-2295. 
When  x—  —  00,  the  signs  are  --  1  ---  ,  giving  2  variations; 


Hence  this  equation  has  but  one  real  root,  and,  consequent 


278  GENERAL  THEORY  OP  EQUATIONS. 

ly,  must  have  two  imaginary  roots.     Moreover  ,%  it  is  easilv 
proved  that  the  real  root  lies  between  0  and  +1. 
Ex.  3.  How  many  real  roots  has  the  equation 


Here  we  have 

X     =       x*-  2x3-  7 
X,    =     4x*-  6x3-14a:+10;  or  2x*-3x*-ftx+5. 
X,,  =   17x*-23x-45 
X,,,  =152x  -305. 
X///y=  +524785. 
When  x=  —  oo,  the  signs  are  +  —  H  ---  h,  giving  4  variations  ; 

x=  +  ct>,  "  +  +  +  +  +,      "       0         " 

Hence  the  four  roots  of  this  equation  are  real. 
If  we  try  different  values  for  x,  we  shall  find  that 
When  #=—3,  the  signs  are  H  ---  1  ---  h,  giving  4  variations; 
x=-2,  "  -  +  +  -+,      "       3 

x=-l,  "  -  +  +  -+,      "       3 

x=     0,  "  +  +  --  +,      "       2 

a?=  +  l,  "  +  ---  +,      "       2 

x=+2,          "  +  ---  +,      "       2 

x=+3,  "  +  +  +++,      "       0 

Hence  this  equation  has  one  negative  root  between  —2  and 
—3  ;  one  negative  root  between  0  and  —1  ;  and  two  positive 
roots  between  2  and  3. 

Ex.  4.  How  many  real  roots  has  the  equation 


Ans.  Three  :  viz.,  two  between  1  and  2,  and  one  between 
—3  and  —4. 

Ex.  5.  How  many  real  roots  has  the  equation 
2x*-20x  +  l9=Ql 

Ans.  Two. 
Ex.  G.  How  many  real  roots  has  the  equation 


Ans.  One  between  1  and  2. 
Ex.  7.  How  many  real  roots  has  the  equation 
x'+3x*+5x-  178=0? 

Ans.  One  between  4  and  5. 


GENERAL  THEORY  OF  EQUATIONS.  279 

Ex.  8.  How  many  real  roots  has  the  equation 
xt-l2x*-\-l2x-3=<)f! 

Ans.  Four. 

Ex.  9.  How  many  real  roots  has  the  equation 
x*-8x3+I4x*+4x+-8=0  ? 

Ans.  Four. 

PROPOSITION  XIV. 

(324.)  To  discover  a  method  of  elimination  for  equations  of 
any  degree. 

The  principle  of  the  greatest  common  divisor  affords  one  of 
the  most  general  methods  for  the  elimination  of  unknown 
quantities  from  a  system  of  equations. 

Suppose  we  have  two  equations  involving  x  and  y  reduced 
to  the  form  of 


If  we  proceed  to  find  the  greatest  common  divisor  of  A  and 
B,  we  shall  have,  according  to  Art.  249, 

A=QB+R. 

But  since  A  and  B  are  each  equal  to  zero,  it  follows  that  R 
must  equal  zero.  Hence  we  see  that,  if  we  divide  one  of  the 
polynomials  by  the  other,  as  in  the  method  of  finding  the 
greatest  common  divisor,  each  successive  remainder  may  be 
put  equal  to  zero.  If  we  arrange  the  polynomials  before  di- 
vision with  reference  to  the  letter  x,  we  shall  at  last  obtain  a 
remainder  which  does  not  contain  x  ;  which  remainder,  being 
put  equal  to  zero,  is  the  equation  from  which  x  has  been  elim- 
inated. 

Ex.  1  .  Eliminate  x  from  the  equations 
a:2+y-13=0, 
x  +y  -   5=0. 

Divide  the  first  polynomial  by  the  second,  as  follows  : 
x'+y*-13  \x+y-5 
x*+(y-5)x\x-y+5 
-(y-5)x+  y'-13 
-(y-5)x- 


2ya—  10y+12  =  remainder. 
13 


280  OUNERAL  THEORY  OP  EQUATIONS. 

This  remainder  we  have  already  proved  must  be  equal  to 
zero  ;  that  is, 


an  equation  from  which  x  has  been  eliminated. 
Ex.  2.  Eliminate  x  from  the  equations 

x*+xy  -56=0, 
xy  +2y*—  60=0. 

Ans.  y4-118y2-f  1800=0. 
Ex.  3.  Eliminate  x  from  the  equations 
.x*+y*—x—  y-78=0, 
xy+x+y—  39=0. 

Ans.  y4+2/3-77t/2-273s/+  1404=0. 
Ex.  4.  Eliminate  x  from  the  equations 
x2—  3xy+y*+y=0, 
x*—xy+l          =0. 

.Arcs.  2/4-5ya+2y-l=0. 

If  we  have  three  equations  containing  three  unknown  quan- 
tities, we  must  first  eliminate  one  of  the  unknown  quantities  by 
combining  either  of  the  equations  with  each  of  the  others.  We 
thus  obtain  two  new  equations  involving  but  two  unknown 
quantities,  from  which  we  may  obtain  a  final  equation  involv- 
ing but  one  unknown  quantity. 

Ex.  5.  Eliminate  x  and  y  from  the  equations 

xyz—  c=0, 
xz+xy+yz—  6=0, 
x  +  y+  z  —  a=0. 

Ans.  z*—  #za+6%  —  c=0. 

Ex.  6.  Eliminate  x  and  y  from  the  equations 
x*+=  7, 


Ans.  z8- 


SECTION  XX. 


SOLUTION  OF  NUMERICAL  EQUATIONS. 

(325.)  We*  will  first  consider  the  method  of  finding  the  in- 
tegral roots  of  an  equation,  and  will  begin  with  forming  the 
equation  whose  roots  are  2,  3,  4,  and  5.  This  equation  must 
be  composed  of  the  factors. 

(x-2)  (x-S)  (x-4)  (x-5)=0. 

If  we  perform  the  multiplication  (which  is  most  expeditious- 
ly  done  by  the  method  of  detached  coefficients  shown  in  Art. 
64),  we  obtain  the  equation 


We  know  that  this  equation  is  divisible  by  x—5.  Let  us 
perform  the  division  by  the  method  of  detached  coefficients 
shown  in  Art.  80. 

A    B     C       D      V         a 

1  —  14+71-154+12011  —  5=  divisor. 

1-5  |l—  9+26-24=  quotient. 

"  9+71 
-  9+45 


+26-154 
+26-130 


-   24+120 
—   24+120. 

Supplying  the  powers  of  x,  we  obtain  for  a  quotient 


This  operation  may  be  still  further  abridged,  as  follows  : 
Represent  the  root  5  by  «,  and  the  coefficients  of  the  given 
equation  by  A,  B,  C,  D,  .....  V, 


282  SOLUTION    OF    NUMERICAL   EQUATIONS. 

We  first  multiply  —a  by  A,  and  subtract  the  product  from 
B;  the  remainder,  —9,  we  multiply  by  —a,  and  subtract  the 
product  from  C ;  the  remainder,  +26,  we  multiply  again  by 
— a,  and  subtract  from  D ;  the  remainder,  —24,  we  multiply 
by  — a,  and,  subtracting  from  V,  nothing  remains.  If  we  take 
the  root  a  with  a  positive  sign,  we  may  substitute  addition  for 
subtraction  in  the  above  statement ;  and  if  we  set  down  only 
the  successive  remainders,  the  work  will  be  as  follows : 

ABC  D  V  a 
I  — 14+71  -154+120[5 
1-  9+26-  24, 

and  the  rule  will  be, 

Multiply  A  by  a,  and  add  the  product  to  B ;  set  down  the  sum, 
multiply  it  by  a,  and  add  the  product  to  C ;  set  down  the  sum, 
multiply  it  by  a,  and  add  the  product  to  D,  and  so  on.  The  final 
product  should  be  equal  to  the  last  term  V,  taken  with  a  contrary 
sign. 

The  coefficients  above  obtained  are  the  coefficients  of  a 
cubic  equation  whose  roots  are  2,  3,  4.  The  equation  may 
therefore  be  divided  by  #—4,  and  the  operation  will  be  as  fol- 
lows : 

l-9+26-24|4 
1-5+6. 

These,  again,  are  the  coefficients  of  a  quadratic  equation 
whose  roots  are  2  and  3.  Dividing  again  by  #—3,  we  have 

l-5+6|3 
1  —  2 
which  are  the  coefficients  of  the  binomial  factor  x—2. 

These  three  operations  of  division  may  be  exhibited  together 
as  follows : 


1-14+71-154+120 
1-  9+26-  24 


5,  first  divisor. 
4,  second  divisor. 
3,  third  divisor. 


1-  5+  6 

1-  2. 

(326.)  The  method  here  explained  will  enable  us  to  find  all 
the  integral  roots  of  an  equation.  For  this  purpose,  we  make 
trial  of  different  numbers  in  succession,  all  of  which  must  be 
divisors  of  the  last  term  of  the  equation.  If  any  division  leaves 


SOLUTION    OF    NUMERICAL    EQUATIONS. 


283 


a  remainder,  we  reject  this  divisor ;  if  the  division  leaves  no 
remainder,  the  divisor  employed  is  a  root  of  the  equation. 
Thus,  by  a  few  trials,  all  the  integral  roots  may  be  easily 
found. 

Ex.  2.  Find  the  seven  roots  of  the  equation 

x^+x*— 14#5—  14#4+49#3+49:2r2— 3Qx— 36=0. 

We  take  the  coefficients  separately,  as  in  the  last  example, 
and  try  in  succession  all  the  divisors  of  36,  both  positive  and 
negative,  rejecting  such  as  leave  a  remainder.  The  operation 
is  as  follows : 

1,  first  divisor. 

2,  second  divisor. 

3,  third  divisor. 

—  1,  fourth  divisor. 

—  I,  fifth  divisor 
—2,  sixth  divisor. 
—3,  seventh  divisor. 


-14-14+49+49-36-36 
1+2-12-26+23+72+36 
1+4—   4—34—45—18 

1+7+17+17+  6 
1+6+11+  6 
1+5+  6 
1+3 
Hence  the  seven  roots  are, 

1,2,3,  -1,  -1,  -2,  -3. 
Ex.  3.  Find  the  six  roots  of  the  equation 

— 1584=0. 

1. 
4. 
6. 

-  2. 

-  3. 
-11. 


1+   5—81—   85+964+  780—1584 
1+  6-75-160+804+1584 
1  +  10-35—300-396 
1  +  16+61+  66 
1  +  14+33 
1  +  11 
The  six  roots,  therefore,  are 

1,4,  6,  -2, -3,  -11. 
Ex.  4.  Find  the  five  roots  of  the  equation 

x*+6x*—10x*—112x*—2Q7x— 110=0. 
1+6-10-112-207-110  -1. 
1+5—15—  97—110  —2. 

1+3-21-   55  -5. 

1-2-11 

Three  of  the  roots,  therefore,  are 
-1,  -2,  -5. 


284  SOLUTION    OF    NUMERICAL    EdUATIONS. 

The  two  remaining  roots  may  be  found  by  the  ordinary 
method  of  quadratic  equations.  Supplying  the  letters  to  the 
last  coefficients,  we  have 

x*-2x-ll=0. 

Hence  x=l±^12. 

Ex.  5.  Find  the  four  roots  of  the  equation 
x*+2x*-1x*-8x+l2=0. 

Ex.  6.  Find  the  four  roots  of  the  equation 
4 


Ex.  7.  Find  all  the  roots  of  the  equation 


Ex.  8.  Find  all  the  roots  of  the  equation 

x6+5x*+x*—  I6x*—  20x—  16=0. 
Ex.  9.  Find  all  the  roots  of  the  equation 


A  -,      ^  J    „ 

Ans.  1,  2,  3,  and  6. 


HORNER'S  METHOD. 

(327.)  The  preceding  method  furnishes  the  roots  of  an  equa- 
tion only  when  they  are  expressed  by  whole  numbers.  When 
the  roots  are  incommensurable,  we  employ  the  following  meth- 
od, which  is  substantially  the  same  as  published  by  Homer  in 
1819. 

The  Theorem  of  Sturm,  together  with  Art.  315,  enables  us 
to  find  the  integral  part  of  any  real  root  of  the  equation  pro- 
posed. We  then  transform  the  equation  into  another  having 
its  roots  less  than  those  of  the  preceding  by  the  number  just 
found,  Art.  316.  We  discover  again,  by  Art.  315,  the  first 
figure  of  the  root  of  this  equation,  which  will  be  the  first  deci- 
mal figure  of  the  root  of  the  original  equation.  Again,  we 
transform  the  last  equation  into  another  having  its  roots  less 
than  those  of  the  preceding  by  this  decimal  figure.  W  e  thus 
discover  the  second  decimal  figure  of  the  root  ;  and  proceed- 
ing in  this  manner  from  one  transformation  to  another,  we  are 
enabled  to  discover  the  successive  figures  of  the  root,  and 
may  carry  the  approximation  to  any  degree  of  accuracy  re- 
quired. 


SOLUTION    OF    NUMERICAL    EaUATION8.  285 

Ex.  1.  Find  a  root  of  the  cubic  equation 


We  have  found,  page  278,  that  this  equation  has  but  one 
real  root,  and  that  it  lies  between  4  and  5.  The  first  figure 
of  the  root,  therefore,  is  4.  To  ascertain  the  second  figure,  we 
transform  the  given  equation  into  another  in  which  the  value 
of  x  is  diminished  by  4,  which  is  done  by  substituting  for  x. 
We  thus  obtain 


The  first  figure  of  the  root  of  this  equation,  according  to 
Art.  315,  is  .5.  Now  transform  the  last  equation  into  another 
in  which  the  value  of  y  is  diminished  by  .5,  which  is  done  by 
substituting  for  ?/,  .54-%.  We  thus  obtain 


The  first  figure  of  the  root  of  this  equation  is  .03.  We  must 
now  transform  this  equation  into  another  in  which  the  value 
of  z  is  diminished  by  .03,  which  is  done  by  substituting  for  z, 
03+u.  We  thus  obtain 

u3+16.59u2+93.7427v=.827623. 

The  first  figure  of  the  root  of  this  equation  is  .008. 

In  order  to  find  the  next  figure,  we  must  transform  the  last 
equation  into  another  in  which  the  value  of  v  is  diminished  by 
.008,  and  so  on. 

(328.)  This  method  would  be  very  laborious  if  we  were 
obliged  to  deduce  the  successive  equations  from  each  other  by 
the  ordinary  method  of  substitution  ;  but  they  may  all  be  de- 
rived from  each  other  by  a  very  simple  law.  Thus,  let 

Ax*+Bx*+Cx=D     (1) 

be  any  cubic  equation  ;  and  let  the  first  figure  of  its  root  be 
denoted  by  «,  the  second  by  a',  the  third  by  a",  and  so  on. 
If  we  substitute  a  for  x  in  equation  (1),  we  shall  have 

Aa3+Ba2+Ca=D,  nearly. 


Whence  a 

If  we  put  y  for  the  sum  of  all  the  figures  of  the  root  except 
the  first,  we  shall  have  x=a+y  ;  and  substituting  this  value 
for  x  in  equation  (1),  we  obtain 


286  SOLUTION    OF    NUMERICAL    EQUATIONS. 


or,  arranging  according  to  the  powers  of  y,  we  have 
Ay»+(B+3A<z)y2+(C+2B«+3Aa>=D-Ca-Ba2-Aa8    (3) 

Let  us  put  B'  for  the  coefficient  of  y2,  C'  for  the  coefficient 
of  y,  and  D'  for  the  right  member  of  the  equation,  and  we  have 

Ay3+B'y2+C'y=D'     (4). 

This  equation  is  of  the  same  form  as  equation  (1)  ;  and,  pro- 
ceeding in  the  same  manner,  we  shall  find 

D' 

a'=C'+B'a'+Aa»     (5)' 

where  a1  is  the  first  figure  of  the  root  of  equation  (4),  or  the 
second  figure  of  the  root  of  equation  (1). 

Putting  %  for  the  sum  of  all  the  remaining  figures,  we  have 
y=a'+z;  and  substituting  this  value  in  equation  (4),  we  shall 
obtain  a  new  equation  of  the  same  form,  which  may  be  written 

Az3+B"z2+C"z=D"     (6)  ; 

and  in  the  same  manner  we  might  proceed  with  the  remaining 
figures. 

Equation  (2)  furnishes  the  value  of  the  first  figure  of  the 
root;  equation  (5)  the  second  figure,  and  similar  equations 
would  furnish  the  remaining  figures.  Each  of  these  expres- 
sions involves  the  unknown  quantity  which  is  sought,  and  might 
therefore  appear  to  be  useless  in  practice.  When,  however, 
the  root  has  been  already  found  to  several  decimal  places,  the 
value  of  the  terms  Ba  and  Aa*  will  be  very  small  compared 

with  C,  and  a  will  be  very  nearly  equal  to  -~.     We  may  there- 

fore employ  C  as  an  approximate  divisor,  which  will  probably 
furnish  a  new  figure  of  the  root.  Thus,  in  the  above  example, 
all  the  figures  of  the  root  after  the  first  are  found  by  division. 

46  -7-77      =.5. 
3.625-7-  92.75=.03. 

.827-7-93.74=.008. 
Tf  we  multiply  the  first  coefficient  A  by  «,  the  first  figure  of 


SOLUTION    OF    NUMERICAL    EQUATIONS.  287 

the  root,  and  add  the  product  to  the  second  coefficient,  we 
shall  have 

B+Aa     (7). 

If  we  multiply  this  expression  by  a,  and  add  the  product  to 
the  third  coefficient,  we  shall  have 

C+B<z+Aaa     (8). 

If  we  multiply  this  expression  by  a,  and  subtract  the  product 
from  D,  we  shall  have 

D-Ca-Baa-Aa3, 
which  is  the  quantity  represented  by  D'  in  equation  (4). 

Again,  multiplying  the  first  coefficient  by  a,  and  adding  the 
product  to  expression  (7),  we  obtain 

B+2Afl     (9). 

Multiplying  this  expression  by  a,  and  adding  the  product  to 
expression  (8),  we  have 


which  is  the  coefficient  of  y  in  equation  (4). 

Again,  multiplying  the  first  coefficient  by  a,  and  adding  the 
product  to  expression  (9),  we  have 

B+3A«, 

which  is  the  coefficient  q/*ya  in  equation  (4). 

We  have  thus  obtained  the  coefficients  of  the  first  trans- 
formed equation  ;  and  by  operating  in  the  same  manner  upon 
these  coefficients,  we  shall  obtain  the  coefficients  of  the  second 
transformed  equation,  and  so  on  ;  and  the  successive  figures 
of  the  root  are  found  by  dividing  D  by  C,  D'  by  C',  D"  by  C", 
and  so  on. 

(329.)  The  preceding  method  is  summed  up  in  the  following 

RULE. 

Represent  the  coefficients  of  the  different  terms  by  A,  B,  C,  and 
the  right-hand  member  of  the  equation  by  D.  Having  found  a, 
the  first  figure  of  the  root,  multiply  A  by  a,  and  add  the  product 
to  B.  Set  down  the  sum  ;  multiply  this  sum  by  a,  and  add  the 
product  to  C.  Set  down  the  sum  ;  multiply  it  by  a,  and  subtract 
the  product  from  D  ;  the  remainder  will  be  the  FIRST  DIVIDEND. 

13* 


288 


SOLUTION    OF    NUMERICAL    EQUATIONS. 


Again,  multiply  A  by  a,  and  add  the  product  to  the  last  num- 
ber under  B.  Multiply  this  sum  by  a,  and  add  the  product  to 
the  last  number  under  C  ;  this  last  sum  will  be  the  FIRST  DIVISOR. 

Again,  multiply  A  by  a,  and  add  the  product  to  the  last  num- 
ber under  B. 

Find  the  second  figure  of  the  root  by  dividing  the  first  divi- 
dend by  the  first  divisor,  and  proceed  with  this  second  figure  pre- 
cisely as  was  done  with  the  first  figure. 

The  second  figure  of  the  root  obtained  by  division  will  fre- 
quently furnish  a  result  too  large  to  be  subtracted  from  the 
remainder  D',  in  which  case  we  must  assume  a  different  figure. 
After  the  second  figure  of  the  root  has  been  obtained,  there 
will  seldom  be  any  further  uncertainty  of  this  kind. 

The  operation  for  finding  a  root  of  the  equation 


will  then  proceed  as  follows  : 


A    B 
1    +3 
4 

7 
4 

C 

+5 
28 
33 
44 

D        a 
=  178     (4.5388=#. 
132 
46  =  1st  dividend. 
42.375 

11 
4 

77  =  1st  divisor.      3.625  =  2d  dividend. 
7.75                          2.797377 

15.5 
.5 

TeTo 

.5 

84.75 
8.00 

.827623  =  3d  dividend. 
.751003872 

92.75  =  2d  divisor.    .076619128  =  4th  dividend. 
.4959 
93.2459 
.4968 

16.53 
3 

16.56 
3 
16.598 

8 

93.  7427=  3d 
.132784 

divisor. 

93.875484 
.132848 

16.606      94.008332  =  4th  divisor. 
Having  found  one  root,  we  may  depress  the  equation 

x*+3x*+5x- 178=0 
to  a  quadratic,  by  dividing  it  by  a— 4.5388.     We  thus  obtain 


SOLUTION    OF    NUMERICAL    EQUATIONS.  289 


where  x  is  evidently  imaginary,  because  q  is  negative  and 
greater  than  ±-.     See  Art.  195. 

After  thus  obtaining  the  root  to  five  or  six  decimal  places, 
several  more  figures  will  be  correctly  obtained  by  simply  di- 
viding the  last  dividend  by  the  last  divisor. 

Ex.  2.  Find  all  the  roots  of  the  equation 


The  first  figure  of  one  of  the  roots  we  readily  find  to  be  3. 
We  then  proceed,  according  to  the  Rule,  to  obtain  the  root  to 
four  decimal  places,  after  which  two  more  will  be  obtained 
correctly  by  division. 

ABC  D       a 

1   +11    -102  =-181     (3.21312=z. 

_3          42  -180 

14      -60  ^1  =  1st  dividend. 

_3          51_  -.992 

17        -^9  =  1st  divisor.        —.008  =  2d  dividend. 
3  4.04  -.006739 

20.2      -4.96  -.001261  =  3d  dividend. 

_  2         4.08  -.001217403 

20.4     -O88  =  2d  divisor.    -.000043597  =  4th  dividend. 
2  .2061 


20.61  -.6739 
1         .2062 

20.62  -  .4677  =  3d  divisor. 
1_       .061899 

20.633  —.405801 
3       .061908 
20.636  -.343893  =  4th  divisor. 

The  two  remaining  roots  may  be  found  in  the  same  way,  or 
by  depressing  the  original  equation  to  a  quadratic.  Those 
roots  are, 

3.22952 
-17.44265. 

When  a  power  of  x  is  wanting  in  the  proposed  equation,  wo 
must  supply  its  place  with  a  cipher. 


290  SOLUTION    OF    NUMERICAL    EQUATIONS. 

Ex.  3.  Find  all  the  roots  of  the  cubic  equation 
z*_7z=_7. 

The  work  of  the  following  example  is  exhibited  in  an  ab- 
breviated form.  Thus,  when  we  multiply  A  by  a,  and  add  the 
product  to  B,  we  set  down  simply  this  result.  We  do  the  same 
in  the  next  column,  thus  dispensing  with  half  the  number  of 
lines  employed  in  the  preceding  example.  Moreover,  we  may 
omit  the  ciphers  on  the  left  of  the  successive  dividends,  if  we 
pay  proper  attention  to  the  local  value  of  the  figures.  Thus, 
it  will  be  seen  that  in  the  operation  for  finding  each  successive 
figure  of  the  root,  the  decimals  under  B  increase  one  place, 
those  under  C  increase  two  places,  and  those  under  D  increase 
three  places. 
1  +0  -7  =-7  (1.356895867=*. 

1  -6  -6 

2  -4=lst  div'r.      —  1=  1st  dividend. 
3.3          -3.01  -.903 

3.6          —  1.93=  2d  div'r.       —97=  2d  dividend. 

3.95         -1.7325  86625 

4.00        -1.5325=  3d  div'r.      10375=  3d  dividend. 

4.056      —1.508164  9048984 

4.062      —1.483792=  4th  div'r.  1326016=  4th  dividend. 

4.0688    —  1  .48053696  1  1  84429568 

4.0696    —1.47728128=  5th  div'r.  141586432=  5th  div'd. 

4.07049  -  1.4769149359  132922344231 

4.07058  -  1.4765485837=  6th  div'r.  8664087769=6th  div'd. 

Having  proceeded  thus  far,  four  more  figures  of  the  root, 
5867,  are  found  by  dividing  the  sixth  dividend  by  the  sixth  di- 
visor. 

We  may  find  the  two  remaining  roots  by  the  same  process  ; 
or,  after  having  obtained  one  root,  we  may  depress  the  equa- 
tion 


to  a  quadratic  equation,  by  dividing  by  #—1.356895867,  and 
we  shall  obtain 

x'+1.356895867a;~5.158833606=0. 
Solving  this  equation,  we  obtain 


x=-.678447933db  V  5.619125204. 


SOLUTION    OF    NUMERICAL    EQUATIONS.  291 


-3.048917, 

Hence  the  three  roots  are         .        .         .          1.356896, 

.692021. 

Ex.  4.  Find  a  root  of  the  equation 


/  -3. 

.        .         .    5      1. 
f      1. 


23  0  =850     (7.0502562208 

17  119  833 

31  336=  1st  divisor.       17=  1st  dividend. 

45.10         338.2550  16.912750 

45.20         340.5150=  2d  divisor.    87250=  2d  dividend. 
45.3004     340.52406008  68104812016 

45.3008     340.53312024=  3d  div.  19145187984=  3d  div'd. 
45.30130  340.5353853050  17026769265250 

45.30140  340.5376503750=4th  div.  21  18418718750=  4th  div. 

Dividing  the  fourth  dividend  by  the  fourth  divisor,  we  ob- 
tain the  figures  62208,  which  make  the  root  correct  to  the 
tenth  decimal  place. 

The  two  remaining  values  of  x  may  be  easily  shown  to  be 
imaginary. 

When  a  negative  root  is  to  be  found,  we  change  the  signs 
of  the  alternate  terms  of  the  equation,  Art.  312,  and  proceed 
as  for  a  positive  root. 

Ex.  5.  Find  a  root  of  the  equation 


Changing  the  signs  of  the  alternate  terms,  it  becomes 


5  +6          +3  +85     (2.16139. 

16  35  70 

26  87=  1st  divisor.         Ts=  1st  dividend. 

36.5         90.65  9.065 

37.0         94.35=  2d  divisor.       5.935=  2d  dividend. 
37.80       96.6180  5.797080 

38.10      98.9040=  3d  divisor.       137920=  3d  dividend. 
38.405    98.942405  98942405 

38.410    98.980815=  4th  divisor.  38977595=  4th  dividend 
38.4165  98.99233995  29697701985 

38.4180  99.00386535=  5th  div'r.     9279893oT5=5th  div'd. 


292  SOLUTION    OF    NUMERICAL    EQUATIONS. 

Hence  one  root  of  the  equation 

5x*-6z*+3z=-Q5 
is  -2.16139. 

The  same  method  is  applicable  to  the  extraction  of  the  cube 
root  of  numbers. 

Ex.  6.  Let  it  be  required  to  extract  the  cube  root  of  9 ;  in 
other  words,  it  is  required  to  find  a  root  of  the  equation 

x*=9. 


I 

0 
2 
4 
6.08 
6.16 
6.24008 
6.24016 

0                               9     (2.0800838. 
4                               8 
12=  1st  divisor.       1=  1st  dividend. 
12.4864                        .998912 
12.9792=  2d  divisor.       1088=  2d  dividend. 
12.9796992064                 1038375936512 

12.9801984192=  3d  d. 

49624063488=  3( 

6.240243  12.980217139929  38940651419787 

6.240246  12.980235860667=  4th  d.  10683412068213=  4th  d. 

Ex.  7.  Find  all  the  roots  of  the  equation 


f  1.02804. 
Ans.   }  6.57653. 
(  7.39543. 
Ex.  8.  Find  all  the  roots  of  the  equation 

?8+9z8+24z+17=0. 

f  -1.12061. 
Ans.   )  -3.34730. 
(  -4.53209. 
Ex.  9.  Extract  the  cube  root  of  48228544. 

Ans.  364. 

Ex.  10.  There  are  two  numbers  whose  difference  is  2,  and 
whose  product,  multiplied  by  their  sum,  makes  120.  What 
are  those  numbers  ? 

Ex.  11.  Find  two  numbers  whose  difference  is  6,  and  such 
that  their  sum,  multiplied  by  the  difference  of  their  cubes,  may 
produce  5040. 

Ex.  12.  There  are  two  numbers  whose  difference  is  4  ;  and 


SOLUTION    OF    NUMERICAL    EClUATIONS.  293 

the  product  of  this  difference,  by  the  sum  of  their  cubes,  is 
3416.     What  are  the  numbers? 

Ex.  13.  Several  persons  form  a  partnership,  and  establish  a 
certain  capital,  to  which  each  contributes  ten  times  as  many 
dollars  as  there  are  persons  in  company.  They  gain  6  plus 
the  number  of  partners  per  cent.,  and  the  whole  profit  is  $392. 
How  many  partners  were  there  ? 

Ex.  14.  There  is  a  number  consisting  of  three  digits  such 
that  the  sum  of  the  first  and  second  is  9 ;  the  sum  of  the  first 
and  third  is  12 ;  and  if  the  product  of  the  three  digits  be  in- 
creased by  38  times  the  first  digit,  the  sum  will  be  336.  Re- 
quired the  number. 

f       636, 

Ans.  <  or  725, 

(  or  814. 

Ex.  15.  A  company  of  merchants  have  a  common  stock  of 
$4775,  and  each  contributes  to  it  twenty-five  times  as  many 
dollars  as  there  are  partners,  with  which  they  gain  as  much 
per  cent,  as  there  are  partners.  Now,  on  dividing  the  profit, 
it  is  found,  after  each  has  received  six  times  as  many  dollars 
as  there  are  persons  in  the  company,  that  there  still  remains 
$126.  Required  the  number  of  merchants. 

Ans.  7,  8,  or  9. 


EQUATIONS  OP  THE  FOURTH  AND  HIGHER  DEGREES. 

(330.)  The  method  already  explained  for  cubic  equations  is 
applicable  to  equations  of  every  degree.  For  the  fourth  de- 
gree, we  shall  have  one  more  column  of  products,  but  the 
operations  are  all  conducted  in  the  same  manner,  as  will  be 
seen  from  the  following  example. 

Ex.  1.  Find  the  four  roots  of  the  equation 


By  Sturm's  Theorem,  we  have  found  that  these  roots  are  all 
real  ;  three  positive,  and  one  negative. 

We  then  proceed  as  follows  : 


294  SOLUTION    OF    NUMERICAL    EQUATIONS. 

1-8+14+4  =8     (5.2360679. 

-3-1-1  -5 

+2+9        +44=  1st  div'r.      13=  1st  dividend. 

7         44  53.288  10.6576 

12.2       46.44      63.072=  2d  div.     2.3424=  2d  dividend. 
12.4       48.92      64.626747  1.93880241 


12.6       51.44      66.193068=3dd.     .40359759=  3d  dividend. 

12.83     51.8249  66.509117736  .39905470641^ 

12.86     52.2107  66.825633024=  4th  d.  4542883584=  4th  div. 

12.89     52.5974 

12.926  52.674956 

12.932  52.752548 

and  by  division  we  obtain  the  four  figures  0679. 

The  other  three  roots  may  be  found  in  the  same  manner. 

{-   .7320508, 
.7639320, 
2  7320508, 
5.2360679. 

Ex.  2.  Find  a  root  of  the  equation 


We  have  found,  by  Sturm's  Theorem,  that  this  equation  has 
a  real  root  between  1  and  2. 

We  then  proceed  as  follows  : 


SOLUTION    OF    NUMERICAL    EdUATIONS. 


295 


co  to 


i—    O    00    C5 

o  en 


en  en  en  en  rf»>  co  to  t— 
^  to 


g  co  to  o  S  K 


o  en  o  O5  co  »— 


§§§ 


en 


O    tfk. 


H-      CO      t—      ^ 

1-1    O    ifk.    H- 


-3    O    O     O 

gggi 


rf^  ^   co  co  H-  ~r 
co  to  oo  en  en  en 


CO 


co  -^  en  o  en  "-1  t-i 

O  CO  O5  tO  00  *5    oa 

to  >-*  q>  ^  en  •-*  «^- 

oo  ^  to  to  ||         P- 

CA?  >^  Cb  f^ 

to  -5  OD  S  to        <. 

o  en  o  05  &       w 


to  to  co        ^. 

if ("  *    1 


& 

o 


>-•  co  en  en  o 

to 

to 

00    H-    00    H 

CO 

en 

to 

to 

00 

Cn 
QD 

to 

ISs-i    c 

Oi 

O5 

CO 

01  'P  ^  a.       to 

1 

en 

3 

CO 

1 

14    1    S 

co  0-        g-        cp 

S 

•^ 

en 

00 

to  &        g         -^ 

1 

en 
II 

CO 

to 

en 

<         r- 

1  ^ 

€-*• 

pi 

zr 

ct> 

296  SOLUTION    OF    NUMERICAL    EQUATIONS. 

Dividing  the  fourth  dividend  by  the  fourth  divisor,  we  ob- 
tain the  figures  789. 

When  we  wish  to  obtain  a  root  correct  to  a  limited  numbei 
of  places,  we  may  save  much  of  the  labor  of  the  operation  by 
cutting  off  all  figures  beyond  a  certain  decimal.  Thus  if,  in 
the  example  above,  we  cut  off  all  beyond  five  decimal  places 
in  the  successive  dividends,  and  all  beyond  four  decimal  places 
in  the  divisors,  it  will  not  affect  the  first  six  decimal  places  in 
the  root. 

Ex.  3.  Find  the  roots  of  the  equation 


C-  3.907378, 

j  +  .443277, 

Ans'   |  +  .606018, 

i.  +2.858083. 

Ex.  4.  Find  the  roots  of  the  equation 

x4-  16z3+79.r2-  140a-=  -58. 

r  +0.58579, 

I  +3.35425, 
Ans.<  +3>41421) 

t  +  8.64575. 
Ex.  5.  Find  the  roots  of  the  equation 


r  +0.934685, 

+3.308424, 

Ans.  J  +3.824325, 

|  +4.879508, 
1^+7.053058. 

Ex.  6.  Required  the  fourth  root  of  18339659776. 

Ans.  368. 

Ex.  7.  Required  the  fifth  root  of  26286674882643. 

Ans.  483. 

Ex.  8.  There  is  a  number  consisting  of  four  digits  such  that 
the  sum  of  the  first  and  second  is  9  ;  the  sum  of  the  first  and 
third  is  10  ;  the  sum  of  the  first  and  fourth  is  11  ;  and  if  the 
product  of  the  four  digits  be  increased  by  36  times  the  product 


SOLUTION    OF    NUMERICAL    EdUATIONS.  297 

of  the  first  and  third,  the  sum  will  be  equal  to  3024  diminished 
by  300  times  the  first  digit.     Required  the  number. 

6345, 
or  7234, 


9012. 


RESOLUTION  OF  EQUATIONS  BY  APPROXIMATION. 

(331.)  The  method  of  Horner  for  finding  the  incommensura- 
ble roots  of  a  numerical  equation  is  generally  better  than  any 
other  ;  nevertheless,  the  method  by  approximation  may  some- 
times be  preferred.  We  shall  explain  the  method  of  Newton, 
and  that  of  Double  Position. 

METHOD  OF  NEWTON. 

This  method  supposes  that  we  have  already  determined 
nearly  the  value  of  one  root  ;  that  we  know,  for  example,  that 
such  a  value  exceeds  «,  and  that  it  is  less  than  a+l.  In  this 
case,  if  we  suppose  the  exact  value  =a+y,  we  are  certain  that 
y  expresses  a  proper  fraction.  Now,  as  y  is  less  than  unity, 
the  square  of  ?/,  its  cube,  and,  in  general,  all  its  higher  powers, 
will  be  much  less  with  respect  to  unity  ;  and  for  this  reason, 
since  we  only  require  an  approximation,  they  may  be  neglect- 
ed in  the  calculation.  When  we  have  nearly  determined  the 
fraction  ?/,  we  shall  know  more  exactly  the  root  a-\-v  ;  from 
which  we  proceed  to  determine  a  new  value  still  more  exact, 
and  we  may  continue  the  approximation  as  far  as  we  please. 

We  will  illustrate  this  method  by  an  easy  example,  requiring 
by  approximation  the  root  of  the  equation 

x*=20. 

Here  we  perceive  that  x  is  greater  than  4,  and  less  than  5. 
If  we  suppose  x=4+y,  we  shall  have 


But,  as  ya  must  be  quite  small,  we  shall  neglect  it,  and  we 
have 

16+8y=20,  or  8^=4. 

Whence  y=.5,  and  x=4.5,  which  already  approaches  near 
the  true  root.     If  we  now  suppose  x=4.5+z,  we  are  sure  thm 


298  SOLUTION    OF    NUMERICAL    EQUATIONS. 

z  expresses  a  fraction  much  smaller  than  y,  and  that  we  may 
neglect  za  with  greater  propriety.     We  have,  therefore, 

a;a=20.25-|-9z=20,  or  9z=— .25. 
Consequently  z=— .0278. 

Therefore,  z=4.5-.0278=4.4722. 

If  we  wish  to  approximate  still  nearer  to  the  true  value,  we 
must  make  x = 4.4722 +v,  and  we  should  have 

za=20.00057284+8.9444u=20. 
So  that  8.9444t>=  —  .00057284. 

Whence  v=  -  .0000640. 

Therefore,  z=4.4722-.0000640=4.4721360, 
a  value  which  is  correct  to  the  last  decimal  place. 

(332.)  The  preceding  method  is  expressed  in  the  following 

RULE. 

Find  by  trial  a  number  (a)  nearly  equal  to  the  root  sought, 
and  represent  the  true  root  by  a+y. 

Substitute  a+y/or  x  in  the  given  equation,  and  there  will  re- 
sult a  new  equation  containing  only  y  and  known  quantities. 

Reject  all  the  terms  of  this  equation  which  contain  the  second 
or  higher  powers  of  y,  and  the  approximate  value  of  y  will  then 
be  given  by  a  simple  equation. 

Havin^  applied  this  correction  to  the  assumed  root,  the  op- 
eration must  be  repeated  with  the  corrected  value  of  a,  when 
a  second  correction  will  be  obtained  which  will  give  a  nearer 
value  of  the  root,  and  the  process  may  be  repeated  as  often  as 
is  thought  necessary. 

EXAMPLES. 
Ex.  1.  Find  a  root  of  the  equation 


If  we  substitute  a +y  for  x  in  this  equation,  and  reject  all  the 
terms  containing  the  higher  powers  of  y,  we  shall  have 


Whence  y= — - 


SOLUTION    OF    NUMERICAL    EaUATIONS.  299 

We  find  by  trial  that  x  is  nearly  equal  to  3.     If  we  substi- 
tute 3  for  «,  we  shall  have 

2 

y=-5r 

Whence  x  =2.  9  nearly. 

And  if  we  substitute  this  new  value  instead  of  a,  we  shall  find 
another  still  more  exact. 

Ex.  2.  Find  a  root  of  the  equation 
x*-6x=IO. 

If  we  make  x=a+y,  we  shall  have 


Therefore,  y 

Assume  <z=2,  and  we  obtain 

5 
y=-—  ,  or  -0.14. 

Hence  #=1.86  nearly. 

If  we  assume  a=1.86,  we  have 

10+11.16-22.262 
y=         59.844-6  -  = 

Hence  x=  1.839  nearly. 

If  we  assume  a=  1.839,  we  shall  have 

_10+11.034-21.033352 
y-  57.18694-6 

Therefore,  x=  1.83901266. 

Ex.  3.  Given  x*—  9#=10,  to  find  one  value  of  x  by  approx- 
imation. 

Ans.  #=3.4494897. 

Ex.  4.  Given  x8+9x9+4.r=80,  to  find  one  value  of  a;  by  ap- 
proximation. 

Ans.  #=2.4721359. 

METHOD  OF  DOUBLE  POSITION. 

(333.)  Another  method  of  finding  the  roots  of  an  equation  is 
by  the  rule  of  Double  Position. 


300  SOLUTION    OF    NUMERICAL    EQUATIONS. 

Substitute  in  the  given  equation  two  numbers  as  near  the 
true  root  as  possible,  and  observe  the  separate  results.  Then 
state  the  following  proportion : 

As  the  difference  of  these  results, 
Is  to  the  difference  of  the  two  assumed  numbers, 
So  is  the  error  of  either  result. 

To  the  correction  required  in  the  corresponding  assumed 
number. 

This  being  added  to  the  number  when  too  small,  or  sub- 
tracted from  it  when  too  great,  will  give  the  true  root  nearly. 
The  number  thus  found,  combined  with  any  other  that  may  be 
supposed  to  approach  still  nearer  to  the  true  root,  may  be  as- 
sumed for  another  operation,  which  may  be  repeated  till  the 
root  is  determined  to  any  degree  of  accuracy  required. 

EXAMPLES. 

Ex.  1.  Given  x*+x*-\-x=WO,  to  find  an  approximate  value 
of  x. 

Having  ascertained,  by  trial,  that  x  is  more  than  4,  and  less 
than  5,  we  substitute  these  two  numbers  in  the  given  equation, 
and  calculate  the  results. 

/    *7*   — — '     4- 

By  the  first  sup-  V    a_16     By-  the  second  sup- 
position, )    s__ft/i         position, 

\  X  —  o4 
Result,  ~84~  Result,  155. 

Then  155-84  :  5-4  : :  100-84  :  .22. 

Therefore,  4 +.22,  or  4.22,  approximates  nearly  to  the  true 
root. 

If,  now,  4.2  and  4.3  be  taken  as  the  assumed  numbers,  and 
substituted  in  the  given  equation,  we  shall  obtain  the  value  of 
x= 4.264  nearly. 

Again,  assuming  4.264  and  4.265,  and  proceeding  in  the 
same  manner,  we  shall  find  x =4.2644299  very  nearly. 

This  rule  is  founded  on  the  supposition  that  the  differences 
in  the  results  are  proportioned  to  the  differences  in  the  as- 
sumed numbers.  This  supposition  is  not  strictly  correct ,  but 
if  we  employ  numbers  near  the  true  values,  the  error  is  gen- 


SOLUTION    OF    NUMERICAL    EQUATIONS.  301 

erally  not  very  great,  and  it  becomes  less  and  less  the  further 
we  carry  the  approximation. 

Ex.  2.  Given  x*+2x*—  23x—109  to  find  one  value  of  x. 

Ans.  x=  5.  13458. 

Ex.  3.  Given  x*—3x2—75x=  10000,  to  find  one  value  of  x. 

Ans.  x=  10.2610. 

Ex.  4.  Given  x6+3x'+2x9—  3x*—  2x=2,  to  find  one  value 
of  x. 

Ans.  x=  1.059109. 

(334.)  We  will  conclude  this  Section  by  finding  some  of  the 
different  roots  of  unity. 

Ex.  1.  Find  the  two  roots  of  the  equation  #a=l,  or  the 
square  roots  of  unity. 

Extracting  the  square  root,  we  find 
x=  +  l,  or  -1. 

Ex.  2.  Find  the  three  roots  of  the  equation  x*=  1,  or  the  cube 
roots  of  unity. 

Since  one  root  of  this  equation  is  x=l9  the  equation  x3—  1=0 
must  be  divisible  by  x—l  ;  and  dividing,  we  obtain 


whence  x  =  —  J=4  ^  ~  3»  or  ~    —  «  -  • 

Hence  the  required  roots  are 

V'^3    -I—  V^3 


+  1, 


which  are  the  cube  roots  of  unity. 

These  results  may  be  easily  verified.  We  have  seen,  on 
page  259,  that  the  cube  of  —  1±V  —  3  is  8,  which,  divided  by 
8,  the  cube  of  the  denominator,  gives  4-1,  as  required. 

Ex.  3.  Find  the  four  roots  of  the  equation  x*=lt  or  the  fourth 
roots  of  unity. 

The  square  root  of  this  equation  is 

**=  +  !,  or  =-1. 
Hence  the  required  roots  are 


802  SOLUTION    OF   NUMERICAL    EQUATIONS. 

+  1,   -1,   +V"^T,  -V=T. 

.Ex.  4.  Find  the  five  roots  of  the  equation  «*=!. 

Since  one  root  of  this  equation  is  x=l,  the  equation  re*— 
must  be  divisible  by  x—  1  ;  and  dividing,  we  obtain 

x'+x*+x*+x+l=0. 
Dividing,  again,  by  a;2,  we  have 

x'+a;+1+I+la=0     (l). 
Now  put  z=x-\  —  . 

X 

Whence  za=xa+2+4 

•€/ 

which,  being  substituted  in  equation  (1),  gives 

za+z_l=0. 
This  equation,  solved  by  the  usual  method,  gives 

«=-i+i>/5i  or  z=-£-iv/5. 
The  values  of  #,  deduced  from  the  equation 


or  x*—  zx=  —  1, 

are 

z         /?^4  z         /za-4 

«=g+  v~T~'  and  ^"s"  V  ~~T"' 

from  which,  by  substituting  the  value  of  z,  we  obtain 


or  =-iIV5+l=Fv'-10+2V5]. 

Hence  the  five  fifth  roots  of  unity  are 
1 


SOLUTION    OF    NUMERICAL    EQUATIONS.  303 

Ex.  5.  Find  the  six  roots  of  the  equation  #6=1. 
These  are  found  by  taking  the  square  roots  of  the  cube  roots. 
Hence  we  have, 


+  1,  -1, 

Thus  we  see  that  unity  has  two  square  roots,  three  cube 
roots,  four  fourth  roots,  jive  fifth  roots,  six  sixth  roots,  and, 
generally,  the  nth  root  of  unity  admits  of  n  different  algebraic 
values.  As,  however,  most  of  these  roots  are  imaginary,  they 
can  not  be  found  by  Horner's  Method. 

14 


.SftdtYA 


npg  adt    m 


;>mfc  &s 

SECTION  XXI. 

;>{^i  o  tear  d*«  edi  <vl 

;^K«ii  »i  ----  -  .  --  •  -- 

.bodlaM  8'ieirioH  y/J  >>nu  :  a  HBO  1 

LOGARITHMS. 

(335.)  In  a  system  of  logarithms,  all  numbers  are  considered 
as  the  powers  of  some  one  number,  arbitrarily  assumed,  which 
is  called  the  base  of  the  system  ;  and  the  exponent  of  that  power 
of  the  base  which  is  equal  to  any  given  number  is  called  the  log- 
arithm of  that  number. 

Thus,  if  a  be  the  base  of  a  system  of  logarithms,  and  <z2=N, 
then  2  is  the  logarithm  of  N  ;  that  is,  2  is  the  exponent  of  the 
power  to  which  the  base  (a)  must  be  raised  to  equal  N. 

If  «3=N/,  then  3  is  the  logarithm  of  N'  for  the  same  reason  ; 
and  if  ax=N",  then  x  is  called  the  logarithm  of  N"  in  the  sys- 
tem whose  base  is  a. 

The  base  of  the  common  system  of  logarithms  (called,  from 
their  inventor,  Briggs'  Logarithms)  is  the  number  10.  Hence 
in  this  system  all  numbers  are  to  be  regarded  as  powers  of  10. 
Thus,  since 

10°=  1,  0  is  the  logarithm  of  1          in  Briggs'  system. 

10^10,         1  "  10 

10a=100,       2  "  100  " 

108=1000,     3  "  1000  " 

104=  10000,  4  "  10000  «' 

&c.,  &c.,  &c. 

From  this  it  appears  that,  in  Briggs'  system,  the  logarithm 
of  every  number  between  1  and  10  is  some  number  between  0 
and  1,  i.  e.,  is  a  proper  fraction.  The  logarithm  of  every  num- 
ber between  10  and  100  is  some  number  between  1  and  2,  i.  e., 
is  1  plus  a  fraction.  The  logarithm  of  every  number  between 


LOGARITHMS.  305 

100  and  1000  is  some  number  between  2  and  3,  i.  e.,  is  2  plus 
a  fraction,  and  so  on. 

(336.)  The  preceding  principles  may  be  extended  to  frac- 
tions by  means  of  negative  exponents.  Thus, 

10"1  or     TV    =0.1  ;        therefore,  —  1  is  the  logarithm  of  .1 

in  Briggs'  system. 

10~2  or    Tio   =0.01  ;  "          -2  "  .01 

10-3or   T¥Vo  =0.001;  "         -3  «  .001 

10-4  or  TO £00  =0.0001  ;          "          -4  "  .0001. 

Hence  it  appears  that  the  logarithm  of  every  number  be- 
tween 1  and  .1  is  some  number  between  0  and  —1,  or  may  be 
represented  by  —  1  plus  a  fraction ;  the  logarithm  of  every 
number  between  .1  and  .01  is  some  number  between  —1  and 
—2,  or  may  be  represented  by  —2  plus  a  fraction;  the  loga- 
rithm of  every  number  between  .01  and  .001  is  some  number 
between  —2  and  —3,  or  is  equal  to  —3  plus  a  fraction,  and  so 
on. 

(337.)  The  logarithms  of  most  numbers,  therefore,  consist 
of  an  integer  and  a  fraction.  The  integral  part  is  called  the 
characteristic,  and  may  always  be  known  from  the  following 

RULE. 

The  characteristic  of  the  logarithm  of  any  number  greater 
than  unity,  is  one  less  than  the  number  of  integral  figures  in  the 
given  number. 

Thus  the  logarithm  of  297  is  2  plus  a  fraction ;  that  is,  the 
characteristic  of  the  logarithm  of  297  is  2,  which  is  one  less 
than  the  number  of  integral  figures.  The  characteristic  of  the 
logarithm  of  5673  is  3  ;  of  73254  is  4,  &c. 

The  characteristic  of  the  logarithm  of  a  decimal  fraction  is  a 
negative  number,  and  is  equal  to  the  number  of  places  by  which 
its  first  significant  figure  is  removed  from  the  place  of  units. 

Thus  the  logarithm  of  .0046  is  3  plus  a  fraction  ;  that  is,  the 
characteristic  of  the  logarithm  is  —3,  the  first  significant  figure, 
4.  being  removed  three  places  from  units. 

In  a  series  of  fractions  continually  decreasing,  the  negative 
logarithms  continually  increase.  Hence,  if  the  fraction  is  in- 
finitely small,  its  logarithm  will  be  infinitely  great ;  that  is,  in 
Briggs'  system,  the  logarithm  of  zero  is  infinite  and  negative. 

U 


306  LOGARITHMS. 

aulq  .  ..  ;.«  &  flsswJsd  ladniufl  amoa  ei  0001  boM  00  i 

GENEEAL  PROPERTIES  OF  LOGARITHMS. 

(338.)  Let  N  and  N'  be  any  two  numbers,  x  and  x1  their  re- 
spective logarithms,  and  a  the  base  of  the  system.  Then,  bv 
the  definition,  Art.  335, 

w   i 

N  =a*      (1). 

Also  N'=ax>     (2). 

Multiplying  together  equations  (1)  and  (2),  we  obtain 

o  m- -•NN'=«rarB/     ^  , 
PMSfifi 

Therefore,  according  to  the  definition  of  logarithms,  a: -fa;'  is 
the  logarithm  of  NN;,  since  x+x'  is  the  exponent  of  that  power 
of  the  base  a  which  is  equal  to  NN' ;  hence 

••.ic 2  a.i  100.  bns  fO.  nawJIsd  •rad'mTm  v*isr?«»  *lo  mHHT 

PROPERTY  I 

The  logarithm  of  the  product  of  two  or  more  factors  is  equal 
to  the  sum  of  the  logarithms  of  those  factors. 

Hence  we  see  that  if  it  is  required  to  multiply  two  or  more 
numbers  by  each  other,  we  have  only  to  add  their  logarithms  ; 
the  sum  will  be  the  logarithm  of  their  product.  We  must  then 
look  in  the  Table  for  the  number  answering  to  that  logarithm, 
in  order  to  obtain  the  required  product. 

EXAMPLES. 

Ex.  1.  Find  the  product  of  8  and  9  by  means  of  logarithms. 

On  page  318,  the  logarithm  of  8  is  given  0.903090 

"  9       "  0.954243 

The  sum  of  these  two  logarithms  is  1.857333, 

which,  according  to  the  same  Table,  is  seen  to  be  the  loga- 
rithm of  72. 

^pla  £-ai  9£00,  lo  mrhn'rap!  erLf  gr/riT 
Ex.  2.  Find  the  continued  product  of  2,  5,  and  14  by  means 

of  logarithms. 

Ex.  3.  Find  the  continued  product  of  1,  2,  3,  4,  and  5  by 
means  of  logarithms. 

(339.)  If,  instead  of  multiplying,  we  divide  equation  (1)  by 
equation  (2),  we  shall  obtain 


LOGARITHMS.  307 


Therefore,  according  to  the  definition,  x—  x'  is  the  logarithm 

N 
of  ^7,  since  x—  x'  is  the  exponent  of  that  power  of  the  base  a 


N 


which  is  equal  to  ^  ;  hence, 


PROPERTY  II. 

The  logarithm  of  a  fraction,  or  of  the  quotient  of  one  number 
divided  by  another,  is  equal  to  the  logarithm  of  the  numerator, 
minus  the  logarithm  of  the  denominator. 

Hence  we  see  that  if  we  wish  to  divide  one  number  by  an- 
other, we  have  only  to  subtract  the  logarithm  of  the  divisor 
from  that  of  the  dividend ;  the  difference  will  be  the  logarithm 
of  their  quotient. 

EXAMPLES. 

Ex.  1.  It  is  required  to  divide  108  by  12  by  means  of  loga- 
rithms. 

The  logarithm  of  108  is  2.033424 

12  1.079181 

The  difference  is  0.954243, 

which  is  the  logarithm  corresponding  to  the  number  9. 

Ex.  2.  Divide  133  by  7  by  means  of  logarithms. 

Ex.  3.  Divide  136  by  17  by  means  of  logarithms. 

Ex.  4.  Divide  135  by  15  by  means  of  logarithms. 

The  preceding  examples  are  designed  to  illustrate  the  prop- 
erties of  logarithms.  In  order  to  exhibit  fully  their  utility  in 
computation,  it  would  be  necessary  to  employ  larger  numbers  ; 
but  that  would  require  a  more  extensive  Table  than  the  one 
given  on  page  318. 

(340.)  Logarithms  are  attended  with  still  greater  advantages 
in  the  involution  of  powers  and  in  the  extraction  of  roots.  For 
if  we  raise  both  members  of  equation  (1)  to  the  mth  power,  we 
obtain 


308  LOGARITHMS. 

Therefore,  according  to  the  definition,  mx  is  the  logarithm  of 
Nm,  since  mx  is  the  exponent  of  that  power  of  the  base  which 
is  equal  to  NM  ;  hence 

PROPERTY  III. 

The  logarithm  of  any  power  of  a  number  is  equal  to  the  loga- 
rithm of  that  number  multiplied  by  the  exponent  of  the  power. 

EXAMPLES. 
Ex.  1  .  Find  the  third  power  of  4  by  means  of  logarithms. 

The  logarithm  of  4  is  0.602060 

Multiply  by  3 

The  product  is  1.806180, 

which  is  the  logarithm  of  64. 

Ex.  2.  Find  the  fourth  power  of  3  by  means  of  logarithms. 

Ex.  3.  Find  the  seventh  power  of  2  by  means  of  loga- 
rithms. 

Ex.  4.  Find  the  third  power  of  5  by  means  of  logarithms. 

(341.)  Also,  if  we  extract  the  mth  root  of  both  members  of 
equation  (1),  we  shall  obtain 


x 
therefore,  according  to  the  definition,  —  is  the  logarithm  of 


Nw;  hence 


PROPERTY  IV. 


The  logarithm  of  any  root  of  a  number  is  equal  to  the  loga- 
rithm of  that  number  divided  by  the  index  of  the  root. 

EXAMPLES. 
Ex.  1.  Find  the  square  root  of  81  by  means  of  logarithms. 

The  logarithm  of  81  is  1.908485 

Divided  by  2 

The  quotient  is  .954243, 

which  is  the  logarithm  of  9. 


LOGARITHMS.  309 

Ex.  2.  Find  the  square  root  of  121  by  means  of  loga- 
rithms. 

Ex.  3.  Find  the  sixth  root  of  64  by  means  of  logarithms. 
Ex.  4.  Find  the  third  root  of  125  by  means  of  logarithms. 

The  preceding  examples  will  suffice  to  show,  that  if  we  had 
tables  which  gave  the  logarithms  of  all  numbers,  they  would 
prove  highly  useful  when  we  have  occasion  to  perform  fre- 
quent multiplications,  divisions,  involutions,  and  extraction  of 
roots. 

(342.)  The  following  examples  will  show  the  application  of 
some  of  the  preceding  principles. 

Ex.  I.  log.  (abcd)=  log.  a+  log.  b+  log.  c-f  log.  d. 

Ex.  2.  log.  f  —  j  =  log.  a+  log.  b+  log.  c—  log.  d—  log.  e. 

Ex.  3.  log.  (ambncp)=m  log.  a+n  log.  b+p  log.  c. 

Ex.  4.  log.  (  —  —\=m  log.  a+n  log.  b—  p  log.  c. 


Ex.  5.  log.  (aa-#a)=log.  [(<*+z)  (a-x)]=  log.  (a+x)  + 
log.  (a-x). 

Ex.  6.  log.  Vrfr=?=i  log.  (a+x)+%  log.  (a-x}. 

Ex.  7.  log.  (<z3V?)=log.  (a?)  =  V  l°g-  a. 

(343.)  We  shall  presently  explain  a  method  by  which  loga- 
rithms may  be  computed.  We  may  observe,  however,  that  it 
is  not  necessary  to  compute  the  logarithms  of  all  numbers  in- 
dependently. From  the  logarithms  of  a  few  numbers,  we  may 
readily  derive  the  logarithms  of  a  great  many  other  numbers. 

We  have  seen,  in  Art.  338,  that  the  logarithm  of  a  product 
is  found  by  adding  together  the  logarithms  of  the  factors.  Let 
us  represent  the  logarithm  of  2  byx;  then,  since  the  logarithm 
of  10  is  1,  we  shall  have 

log.  20     =x+l,         log.  20000     = 
log.  2000=x+3,         log.  2000000=,  &c. 

We  have  seen,  in  Art.  340,  that  the  logarithm  of  any  power 
of  a  number  is  equal  to  the  logarithm  of  that  number  multiplied 
bv  the  exponent  of  the  power. 


310  LOGARITHMS. 

Hence,  log.  4  =2z,        log.  32  = 

log.  16=4#,        log.  128=,  &c. 

Hence  we  find,  also,  that 

log.  40     =2#+l,  log.  4000     = 

log.  400  =2z+2,  log.  40000  =,  &c.     q  9Jy 

log.  80     =3z+l,  log.  8000     = 

log.  800  =3x+2,  log.  80000  =,  &c. 

log.  160  =4x+l,  log.  16000  = 

log.  1600=4^+2,  log.  160000=,  &c. 

We  have  seen,  in  Art.  339,  that  the  logarithm  of  a  fraction 
is  equal  to  the  logarithm  of  the  numerator  minus  the  logarithm 
of  the  denominator.  Hence,  log.  5=  log.  (y)  =  l— #. 

Hence,  log.  50      =2—  x,  log.  5000      = 

log.  500  =3—  x,  log.  50000  =,  &c. 

log.  25   =2— 2x,  log.  625 

log.  125  =3-3#,  log.  3125   =,  &c. 

log.  250  =3-2z,  log.  25000  = 

log.  2500  =4— 2o;,  log.  250000  =,  &c. 

log.  1250  =4— 3#,  log.  125000  = 

log.  12500=5-3*,  log.  1250000=,  &c. 

log.  6250  =5-4;r,  log.  625000  = 

log.  62500=6— 4x,  log.  6250000=,  &c. 

(343.)  So,  also,  from  the  logarithm  of  3  we  might  easily  de- 
rive a  great  number  of  other  logarithms.  From  the  Table  on 
page  318,  we  find  the  logarithm  of  3  to  be  .477121 :  it  is  re- 
quired to  derive  from  this  the  logarithm  of  30. 

Required  the  logarithm  of  3000. 

Required  the  logarithm  of  9. 

Required  the  logarithm  of  27. 

Required  the  logarithm  of  81. 

Required  the  logarithm  of  90. 

Required  the  logarithm  of  270. 

Required  the  logarithm  of  900. 

From  the  same  Table,  we  find  the  logarithm  of  2  to  be 
.301030  :  it  is  required,  by  the  aid  of  the  logarithms  of  3  and 
2,  to  obtain  the  logarithm  of  6. 

Required  the  logarithm  of  12. 


LOGARITHMS.  311 

Required  the  logarithm  of  15. 
Required  the  logarithm  of  18. 

From  the  same  Table,  we  find  the  logarithm  of  5  to  be 
.698970.  It  is  required  from  this  to  deduce  the  logarithm 
of  50. 

Required  the  logarithm  of  500. 
Required  the  logarithm  of  5000. 

From  the  same  Table,  we  find  the  logarithm  of  95  to  be 
1.977724.  The  logarithm  of  9.5,  or  f f,  is  equal  to  the  loga- 
rithm of  95  minus  the  logarithm  of  10. 

Hence  the  logarithm  of  9.5  is  0.977724. 
Also,  the  logarithm  of  950  is     2.977724. 

Hence  the  decimal  part  of  the  logarithm  of  any  number  is  the 
same  as  that  of  the  number  multiplied  or  divided  by  10,  100, 
1000,  &c. 

Prime  numbers  are  such  as  can  not  be  decomposed  into  fac- 
tors ;  as,  2,  3,  5,  7,  11,  13,  17,  &c.  All  other  numbers  arise 
from  the  multiplication  of  prime  numbers.  If,  therefore,  we 
knew  the  logarithms  of  all  the  prime  numbers,  we  could  find 
the  logarithms  of  all  other  numbers  by  simple  addition. 

(345.)  We  will  now  explain  a  method  by  which  the  loga- 
rithm of  any  number  may  be  computed. 

If  a  series  of  numbers  be  taken  in  Geometrical  progression, 
their  logarithms  will  form  a  series  in  Arithmetical  progression. 
Thus,  take  the  geometrical  series 

1,  10,  100,  1000,  10000,  100000, 

their  logarithms  are 

0,  1,  2,  3,  4,  5, 
forming  an  arithmetical  series. 

If,  now,  we  find  a  geometrical  mean  between  any  two  num- 
bers in  the  first  series,  its  logarithm  will  be  the  arithmetical 
mean  between  the  two  corresponding  numbers  in  the  lower 
series. 

Find,  for  example,  a  geometrical  mean  between  1  and  10. 
It  will  be  the  square  root  of  10,  or  3.162277.  The  arithmet- 
ical mean  between  0  and  1  is  0.5. 


312  LOGARITHMS. 

Therefore,  the  logarithm  of  3.162277  is  0.5. 

Find,  again,  a  geometrical  mean  between  3.162277  and  10, 
which  is  5.623413.  Find,  also,  the  arithmetical  mean  between 
0.5  and  1,  which  is  0.75.  .  { 

Therefore,  the  logarithm  of  5.623413  is  0.75. 

Find,  now,  a  geometrical  mean  between  3.162277  and 
5.623413,  which  is  4.216964.  Its  logarithm  will  be  the  arith- 
metical mean  between  0.5  and  0.75,  which  is  0.625. 

Therefore,  the  logarithm  of  4.216964  is  0.625. 

Find,  again,  a  geometrical  mean  between  4.216964  and 
5.623413,  which  is  4.869674.  Its  logarithm  will  be  the  arith- 
metical mean  between  0.625  and  0.75,  which  is  0.6875. 

Thus  we  have  found  the  logarithms  of  four  new  numbers, 
and  in  this  manner  we  might  proceed  to  construct  a  table  of 
logarithms.  It  will  be  observed  that  these  numbers  are  all 
fractional,  whereas  it  is  most  convenient  to  have  the  loga- 
rithms of  integers.  By  pursuing  this  method,  however,  we 
might  eventually  find  the  logarithm  of  a  whole  number ;  as, 
for  example,  5.  For  we  have  already  found  the  logarithm  of 

5.623413  to  be  0.75, 
and  the  logarithm  of  4.869674     "      0.6875. 

One  of  these  numbers  is  greater  than  5,  and  the  other  less. 
A  geometrical  mean  between  them  is  5.232991,  which  is  too 
great ;  but  the  mean  between  this  result  and  the  last  of  the  two 
preceding  is  5.048065,  which  is  already  a  close  approximation. 
By  pursuing  the  same  method,  we  may  come  nearer  and  near- 
er to  the  number  5,  until  at  last,  after  finding  twenty-two  geo- 
metrical means,  the  difference  is  inappreciable  in  the  sixth 
decimal  place,  and  we  obtain 
the  logarithm  of  5  equal  to  0.69897 ; 
and,  by  a  like  process,  the  logarithm  of  any  other  number  may 
be  found. 

(346.)  Hence,  to  compute  the  logarithm  of  any  number,  we 
have  the  following 

RULE. 

Take  the  geometrical  series  1, 10, 100, 1000, 10000,&c.,  and  ap- 
ply to  it  the  arithmetical  series  0,  1,2,  3,  4,  &c.,  as  logarithms. 


LOGARITHMS.  313 

Find  a  geometrical  mean  between  1  and  10,  10  and  100,  or 
any  other  two  terms  of  the  first  series  between  which  the  proposed 
number  lies. 

Between  the  mean  thus  found  and  the  nearest  term  of  the  first 
series,  find  another  geometrical  mean  in  the  same  manner,  and 
so  on,  till  you  approach  as  near  as  is  necessary  to  the  number 
whose  logarithm  is  sought. 

Find,  also,  as  many  arithmetical  means  between  the  correspond- 
ing terms  0,  1,  2,  3,  4,  &c.,  of  the  other  series,  in  the  same  order 
as  the  geometrical  ones  were  found ;  the  last  of  these  will  be  the 
logarithm  answering  to  the  number  required. 

In  this  manner  were  the  logarithms  of  all  the  prime  num- 
bers at  first  computed  ;  but  much  more  expeditious  methods 
have  since  been  devised. 

Having  obtained  the  logarithm  of  5,  it  is  easy  to  find  the 
logarithm  of  2.  For  the  logarithm  of  2=  log.  (V)=  l°g-  10~ 
log.  5=1-0.69897=0.30103. 

LOGARITHMIC  SERIES. 

(347.)  The  preceding  method  of  computing  logarithms  is 
very  laborious  in  practice.  It  is  found  much  more  convenient 
to  express  the  logarithm  of  a  number  in  the  form  of  a  series. 

Let  x  be  a  number  whose  logarithm  is  required  to  be  de- 
veloped in  a  series,  and  let  us  employ  the  method  of  Unknown 
Coefficients.     It  is  plain  that  we  can  not  assume 
log.  x=A+Rx+Cx*+,  &c.  ; 

for  when  we  make  x=Q,  the  first  member  reduces  to  infinity, 
while  the  second  member  reduces  to  A,  a  finite  quantity. 
Neither  can  we  suppose 

log.  x= Ax+Bx*+Cx*+,  &c. : 
for  when  we  make  x=Q,  we  have 

log.  0  (which  is  infinite),  equal  to  zero, 
which  is  absurd. 
But  if  we  suppose 

log.  (l+x)  =  Ax+Bx'+Cz'+'Dx*+,  &c.     (1), 
when  we  make  x=0,  the  equation  becomes 

log.  1,  equal  to  zero, 
which  is  conformable  to  Art.  335. 


314  LOGARITHMS. 

Let  us  also  assume 

log.  (l+z)=A%+Bz'+C 
Subtracting  equation  (2)  from  (1),  we  obtain 
log.  (l+^-bg.  (i+z)=A(x-z)+E(x*-z')+C(x*-z*)+9 


&c.     (3). 

The  second  member  of  this  equation  is  divisible  by  x—z, 
Art.  76  ;  we  will  reduce  the  first  member  to  a  form  in  which 
it  shall  also  be  divisible. 


We  have  log.  (1+*)-  log.  (!+«)=  log.  (-^  )  = 

id 
—  1 


log.  (i+y—-; 


Now,  since  —  —  may  be  regarded  as  a  single  quantity,  v,  we 

may  develop  log.  (1+v)  in  the  same  manner  as  log.  (1+x), 
which  gives 


This  last  series  must  be  identical  with  the  one  which  we 
have  already  obtained  for  log.  (  1+-  -  j,  or  its  equal,  log. 

(!+#)—  log.  (1+z),  in  equation  (3)  ;  and  since  the  terms  of 
both  are  divisible  by  x~-  z,  by  canceling  this  common  factor, 
we  obtain 

—zY 


+  Z3)  +  ,  &C. 

Since  this  equation,  like  the  preceding,  must  be  verified  for 
all  values  of  x  and  z,  the  equality  must  subsist  when  x=z.  But 
on  this  hypothesis,  all  the  terms  of  the  first  series  vanish  ex- 
oept  one,  and  we  have 

-,  &c.  ; 

or,  performing  the  division  indicated  in  the  first  member,  we 
obtain 


LOGARITHMS.  315 

Therefore,  according  to  the  principle  of  Art.  302,  we  have 
the  equations 

A=A, 

— A=2B;  whence  B=— — . 


-A=4D;       «       D=~T 

A=5E;        «       E==+:f- 

The  law  of  the  series  is  obvious  ;  and  hence,  substituting  the 
values  of  B,  C,  D,  &c.,  in  equation  (1),  we  obtain,  for  the  de- 
velopment of  log. 


log.  ^T-*,,--.. 

1                   &                *J  -X 

(^y»  xy*2          «3  ™4          _^B          ^,6                         \ 

*U  JU            %JU  JU            *JU            JU                             \ 


The  number  A  is  called  the  modulus  of  the  system  of  loga- 
rithms employed.  Lord  Napier,  the  illustrious  inventor  of  log- 
arithms, assumed  the  modulus  equal  to  unity.  If,  then,  we  des- 
ignate Naperian  logarithms  by  log.',  we  shall  have 

log.'(l+-)=f-|+|-J+f-|V,&c.    (4). 

By  giving  to  x  in  succession  all  possible  values,  we  may  ob- 
tain from  this  equation  the  logarithms  of  all  numbers. 

If  we  make  x=0,  we  shall  have  log.'  1=0. 

Make  x=l,  and  we  obtain 

log.'  2=l-i+i-i+i-»  &c., 

a  series  which  converges  so  slowly  that  it  would  be  necessary 
to  employ  a  very  large  number  of  terms  to  obtain  the  accuracy 
desirable.  The  series  may  be  rendered  more  converging  in 
the  following  manner : 

In  equation  (4),  substitute  —x  for  x,  and  it  becomes 

log.' (l-*)  =  -f- 1-|-~ ,  &c.     (5). 


316  LOGARITHMS. 

Subtracting  equation  (5)  from  equation  (4),  and  observing 

14-  x 

that  log.'  (I+x)-  log.'  (l-z)=  log.'r-1—  ,  we  obtain 

1  ~~~x 


x 


Put  a;==or»  an(*  ^e  Prece(iing  series  becomes,  by  substi- 


tution, 
log/  -  log.'  (*+!)-  log,  Z= 

5(2z+l)6+"  • 

(348.)  The  last  series  may  be  employed  for  computing  the 
logarithm  of  any  number,  when  the  logarithm  of  the  preceding 
number  is  known.  Making  successively  2=1,  2,  4,  6,  &c.,  we 
obtain  the  following 

NAPERIAN,  OB  HYPERBOLIC  LOGARITHMS. 

log.'    2=2(|  +T^+7^+TT?+  •  •  -  )  =  0.693147 

log.'    3=  log.'  2+2(^^+7^+-^+  '  i  •?         =  L098612 
log.'    4=2  log.'  2  =1.386294 

log.'    5=10^4+2(7+-^+^+^+  .  .  .)          =  1.609438 
log.'    6=  log.'  3+  log.'  2  =1.791759 

log.'    7=  log.'  6+2(^+7^+77^+-^+  .  .  .  )=  1.945910 
log.'    8=  3  log.'  2  =  2.079442 

log.'    9=2  log.'  3  =2.197225 

log.'  10=  log.'  5+  log.'  2  =  2.302585 

&c.,  &c.,  &c. 

(349.)  The  Naperian  logarithms  being  computed,  it  is  easy 
to  form  any  other  system.  We  have  found 

x    a:9    z8    z4    x6    x6  \ 


Distinguishing  the  Naperian  logarithms  by  an  accent,  we 
have 


LOGARITHMS.  317 

Hence 

log.  (l+x)  :  log.'  (l+x)  :  :  A  :  A'. 

Therefore,  the  logarithms  of  the  same  number  in  different  sys- 
tems are  to  each  other  as  the  moduli. 

In  Napier's  system,  the  modulus  =1.     Hence 
log.  (!+*)=  A.  log.'  (l+x). 

That  is,  the  common  logarithm  of  a  number  is  equal  to  its 
Naperian  logarithm  multiplied  by  the  modulus  of  the  common 
system. 

If,  then,  we  knew  the  modulus  of  the  common  system,  we 
could  easily  convert  the  preceding  Naperian  logarithms  into 
common  logarithms.  Now,  from  the  equation 

log.  (!+#)  =  A.  log'.  (l+x),  we  obtain 
log,   (l+x) 
log.' 


log.   10 
Suppose  x=9,  then  A=  ]Qg/  1Q. 

But  log.  10=1.     Hence 


which  is  the  modulus  of  the  common  system. 
(350.)  We  can  now  compute  the 

COMMON,  OB  BRIGGS'  LOGARITHMS. 

log.    2=0.693147X0.434294  =0.301030 

log.    3=1.098612X0.434294  =0.477121 

log.    4=2  log.  2  =0.602060 
log.    5=  log.  10-  log.  2=1  -  log.  2=0.698970 

log.    6=  log.  3+  log.  2  =0.778151 

log.    7=1.945910X0.434294  =0.845098 

log.    8=3  log.  2  =0.903090 

log.    9=2  log.  3.  =0.954243 

log.  10=  =1.000000 

&c.,         &c.,  &c. 


318 


LOGARITHMS. 


We  thus  obtain  the  following  Table  of  Common  Logarithms 


No. 

Logarithm. 

No. 

Logarithm. 

No. 

Logarithm. 

No. 

Logarithm. 

1 

2 
3 
4 

c 
o 

0.000000 
0.301030 
0.477121 
0.602060 
0.698970 

36 
37 
38 
39 
40 

1.556303 
1.568202 
1.579784 
1.591065 
1.602060 

71 
72 
73 
74 
75 

1.851258 
1.857332 
1.863323 
1.869232 
1.875061 

106 
107 
108 
109 
110 

2:025306 
2.029384 
2.033424 
2.037426 
2.041393 

6 

°7 
8 
9 
10 

0.778151 
0.845098 
0.903090 
0.954243 
1.000000 

41 
42 
43 
44 
45 

1.612784 
1.623249 
1.633468 
1.643453 
1.653213 

76 

77 
78 
79 
80 

1.880814 
1.886491 
1.892095 
1.897627 
1.903090 

111 
112 
113 
114 
115 

2.045323 
2.049218 
2.053078 
2.056905 
2.060698 

11 
12 
13 
14 
15 

1.041393 
1.079181 
1.113943 
1.146128 
1.176091 

46 
47 
48 
49 
50 

1.662758 
1.672098 
1.681241 
1.690196 
1.698970 

81 
82 
83 

84 
85 

1.908485 
1.913814 
1.919078 
1.924279 
1.929419 

116 
117 
118 
119 
120 

2.064458 
2.068186 
2.071882 
2.075547 
2.079181 

16 
17 
18 
19 
20 

1.204120 
1.230449 
1.255273 
1.278754 
1.301030 

51 
52 
53 
54 
55 

1.707570 
1.716003 
1.724276 
1.732394 
1.740363 

86 
87 
88 
89 
90 

1.934498 
1.939519 
1.944483 
1.949390 
1.954243 

121 
122 
123 
124 
125 

2.082785 
2.086360 
2.089905 
2.093422 
2.096910 

21 
22 
23 
24 
25 

1.322219 
1.342423 
1.361728 
1.380211 
1.397940 

56 
57 

58 
59 
60 

1.748188 
1.755875 
1.763428 
1.770852 
1.778151 

91 
92 
93 
94 
95 

1.959041 
1.963788 
1.968483 
1.973128 
1.977724 

126 
127 
128 
129 
130 

2.100371 
2.103804 
2.107210 
2.110590 
2.113943 

26 
27 

28 
29 
30 

1.414973 
1.431364 
1.447158 
1.462398 
1.477121 

61 
62 
63 
64 
65 

1.785330 
1.792392 
1.799341 
1.806180 
1.812913 

96 
97 
98 
99 
100 

1.982271 
1.986772 
1.991226 
1.995635 
2.000000 

131 
132 
133 
134 
135 

2.117271 
2.120574 
2.123852 
2.127105 
2.130334 

31 
32 
33 
34 
35 

1.491362 
1.505150 
1.518514 
1.531479 
1.544068 

66 
67 
68 
69 
70 

1.819544 
1.826075 
1.832509 
1.838849 
1.845098 

101 
102 
103 
104 
105 

2.004321 
2.008600 
2.012837 
2.017033 
2.021189 

136 
137 
138 
139 
140 

2.133539 
2.136721 
2.139879 
2.143015 
2.146128 

(351.)  Let  us  now  determine  the  base  of  Napier's  system. 
Designating  it  by  a,  we  shall  have,  Art.  349, 

log/  a  :  log.  a  :  :  1  :  0.434294. 
But  log/  a=l.     Hence 

log.  a=0.434294. 


LOGARITHMS.  319 

That  is,  the  modulus  of  the  common  system  is  equal  to  the 
common  logarithm  of  Napier's  base. 

We  wish,  then,  to  find  the  number  corresponding  to  the 
common  logarithm  0.434294.  By  inspecting  the  preceding 
table,  we  see  that  this  number  must  be  a  little  less  than  3. 
More  accurately,  it  is 

2.718282, 

which  is  the  base  of  Napier's  system. 

Any  number,  except  unity,  may  be  taken  as  the  base  oi  a 
system  of  logarithms,  and  hence  there  may  be  an  infinite  num- 
ber of  systems.  Only  two  systems,  however,  are  much  used ; 
those  of  Briggs  and  Napier. 

The  base  of         Briggs'  system  is  10. 

Napier's        "          2.718282. 
The  modulus  of  Briggs'          "          0.434294. 
"  Napier's        "          1. 

Hence,  in  Briggs'  system,  all  numbers  are  to  be  regarded  as 
powers  of  10. 

Thus,  100801=2, 

1004"=3, 

10°-8oa=4, 
10°-698=5, 
&c.,  &c. 

In  Napier's  system,  all  numbers  are  to  be  regarded  as  pow- 
ers of  2.718282. 

Thus,  2.718°-698=2, 

2.7181-098=3, 

2.7181'886=4, 

2.718'-608=5, 

&c.,     &c. 

flriggs'  logarithms  are  employed  in  all  the  common  opera- 
tions of  multiplication  and  division,  and  hence  they  are  known 
by  the  name  of  common  logarithms.  Napier's  logarithms  are 
of  great  use  in  the  application  of  the  calculus  to  many  analyt- 
ical and  physical  problems.  They  are  also  called  hyperbolic 
logarithms,  having  been  originally  derived  from  the  hyperbola. 


•320  LOGARITHMS. 


EXPONENTIAL  EQUATIONS. 

(352.)  An  exponential  quantity  is  one  which  is  raised  to 
some  unknown  power,  or  which  has  an  unknown  quantity  for 
an  exponent  ;  as, 

i  i 

ax,  ax,  xx,  or  x*,  &c. 

An  exponential  equation  is  one  which  contains  an  exponen- 
tial quantity  ;  as, 

a*=b,  x*=c,  &c. 

Such  equations  are7  most  easily  solved  by  means  of  loga- 
rithms. Thus,  consider  the  equation 

ax=b. 

Taking  the  logarithm  of  each  member  of  the  equation,  we 
have 

x  log.  a=  log.  6, 

log.  b 

or  x=  r-5  —  . 
log.  a 

Ex.  1.  What  is  the  value  of  x  in  the  equation  3*=81  ? 
By  the  preceding  formula,  x—  ,          . 

Looking  out  the  logarithms  of  81  and  3  from  the  Table  on 
page  318,  we  have 

_1.908485_ 
=  .477121" 

Therefore,  34=81. 

Ex.  2.  What  is  the  value  of  a;  in  the  equation  3*=20? 
log.  20     1.301030 


Therefore,  33'7"=20  nearly. 

Ex.  3.  What  is  the  value  of  a;  in  the  equation  5*=  12? 


(2\  x    3 
3/   =4 

(353.)  The  other  equation,  af=c,  may  be  solved  by  trial  as 


LOGARITHMS.  321 

in  Art.  333.     Thus,  taking  the  logarithm  of  each  member,  we 
have 

x  log.  x=  log.  c. 

Find  now,  by  trial,  two  numbers  nearly  equal  to  the  value 
of  x,  and  substitute  them  for  x  in  the  given  equation.  Then 
say, 

As  the  difference  of  these  results, 

Is  to  the  difference  of  the  two  assumed  numbers, 

So  is  the  error  of  either  result, 

To  the  correction  required  in  the  corresponding  assumed 
number. 

Ex.  1.  Given  0^=100  to  find  the  value  of  x. 

Here  we  have      x  log.  x=  log.  100=2. 

Suppose  x=3, 

then          0.477121  X  3=  1.431363,  which  is  too  small. 

Suppose  x=4, 

then          0.602060X4=2.408240,  which  is  too  great. 

Hence  the  value  of  x  is  between  3  and  4,  but  nearer  to  4. 
Assume,  then,  3.5  and  3.6  for  the  two  numbers. 

By  the  first  supposition,  By  the  second  supposition, 
x=3.5',  log.  x=  .544068  a;=3.6;  log.  x—  .556303 
Multiplied  by  3.5  Multiplied  by  3.6 

x  .  log.  x=  1.904238  x .  log.  x=2.002689 

Diff.  of  results  :  Diff.  assumed  numbers  :  :  Error  of  2d  result  :  Its  correction. 
.098451     :  0.1  ::       .002689        :     .00273 

Hence          z=3.6-.00273=3.59727  nearly. 
Therefore,         3.59727369m=100  nearly. 

If  we  wish  a  more  accurate  result,  the  operation  must  be  re- 
peated with  two  new  numbers ;  as,  for  example,  3.59727  and 
3.59728. 

Ex.  2.  Given  xx=Q,  to  find  the  value  of  x. 

Ex.  3.  Given  xx=20x,  to  find  an  approximate  value  of  x. 

COMPOUND  INTEREST. 

(354.)  In  calculating  compound  interest,  the  first  subject  of 
inquiry  is,  to  what  sum  does  a  given  principal  amount,  after  a 


322  LOGARITHMS. 

certain  number  of  years,  the  interest  being  annually  added  to 
the  principal?  It  is  evident  that  $1.00,  placed  out  at  5  per 
cent.,  becomes,  at  the  end  of  a  year,  a  principal  of  $1.05.  But 
the  amount  at  the  end  of  each  year  must  be  proportioned  to 
the  principal  at  the  beginning  of  the  year.  In  order,  then,  to 
find  the  amount  at  the  end  of  two  years,  we  institute  the  pro- 
portion 

1.00  :  1.05  :  :  1.05  :  (1.05)a. 

The  sum  1.059  must  now  be  considered  as  the  principal,  and 
hence,  to  find  the  amount  at  the  end  of  three  years,  we  say 

1.00  :  1.05  :  :  (1.05)2  :  (1.05)3. 

And  in  the  same  manner  we  find  that  the  amount  of  $1.00 
for  n  years  at  compound  interest  is  (1.05)ro. 

If  the  rate  of  interest  were  six  per  cent.,  we  should  find  the 
amount  for  n  years  to  be  (1.06)n. 

Th6  amount  of  two  dollars  for  a  given  time  must  obviously 
be  double  the  amount  of  one  dollar,  and  the  amount  of  $1000 
must  be  a  thousand  times  the  amount  of  one  dollar. 
Hence,  if  we  put  P  to  represent  the  principal, 

r  the  rate  per  cent,  considered  as  a  decimal, 
n  the  number  of  years, 
A  the  amount  of  the  given  principal  for  n 
years,  we  shall  have 

A=P.(l+r)". 

This  equation  contains  four  quantities,  A,  P,  w,  r ;  any  three 
of  which  being  given,  the  fourth  may  be  found.  The  computa- 
tions are  most  readily  performed  by  means  of  logarithms. 
Taking  the  logarithms  of  both  members  of  the  preceding  equa- 
tion, and  reducing,  we  find 

l.log.A        =»Xlog.(l+r)+log.P 

2.  log.  P     .,:£=log.  A-nX  log.  (l+r^f 

log.  A-  log.  P 

3.  log.  (l+r)=— 2 . 


. .,,   .  .  / 


log.  A-  log.  P 
log.  (1+r)     ' 

EXAMPLES. 

Ex.  1.  What  is  the  amount  of  twenty  dollars,  at  6  per  cent. 
compound  interest,  for  11  years? 


LOGARITHMS.  323 

Jn  this  example  we  employ  formula  (1). 

Amount  of  $1.00  for  1  year  $1.06,      log.  =0.025306 

Multiplying  by  11,  H. 

0.278366 

Given  principal  820.  log.  =1.301030 

Amount  $38  nearly,  1.579396. 

This  result  is  derived  from  the  Table  on  page  318.     By  con- 
sulting a  larger  Table,  we  should  find  the  amount  $37.97. 

Ex.  2.  What  principal  at  5  per  cent,  interest  will  amount  to 
$66  in  13  years? 

Here  we  employ  formula  (2). 

l+r=1.05,  log.  =0.021189 

Multiplying  by  n,  13 

Subtract  0.275457 

From  log.  A,  1.819544 

P=$35  nearly,  1.544087. 

Ex.  3.  At  what  rate  per  cent,  must  $40  be  put  out  at  com- 
pound interest,  that  it  may  amount  to  $57  in  9  years  ? 
Here  we  employ  formula  (3). 

A=57,  log.  =1.755875 

P  =40,  log.  =1.602060 

Dividing  by  n,  9)0.153815 

l+r=1.04  =0.017091 

Consequently,  r=.04,  or  four  per  cent. 

How  could  this  result  be  obtained  without  the  use  of  loga- 
rithms ? 

Ex.  4.  In  what  time  will  $50  amount  to  $90  at  5  per  ce$t. 
Here  we  employ  formula  (4). 

A=90,  log,  =1.954243 

P  =  50,  log.  =1.698970 

1  +r=  1.05,  whose  logarithm  is  0.021189)0.255273. 

Dividing  one  logarithm  by  the  other,  we  obtain  12,  Ans. 

Ex.  5.  What  is  the  amount  of  $52  at  3  per  cent,  compound 
interest  for  15  years? 

Ans.  $81. 


324  LOGARITHMS. 

Ex.  6.  What  principal  at  6  per  cent,  compound  interest  will 
amount  to  $101  in  4  years? 

Ans.  880. 

Ex.  7.  At  what  rate  will  810  amount  to  $16  in  16  years? 

Ans.  Three  per  cent. 

Ex.  8.  What  will  8300  amount  to  in  10  years  at  compound 
interest  semi-annually,  the  yearly  rate  being  6  per  cent.  ? 

Ex.  9.  In  what  time  will  a  sum  of  money  double  at  6  per 
cent,  compound  interest  ? 

Ans.  11.89  years. 

Ex.  10.  In  what  time  will  a  sum  of  money  triple  itself  at  4 
per  cent,  compound  interest  ? 

Ans.  28.01  years. 

(355.)  The  natural  increase  of  population  in  a  country  may 
be  computed  in  the  same  way  as  compound  interest.  Know- 
ing the  population  at  two  different  dates,  we  compute  the  rate 
of  increase  by  formula  (3),  and  from  this  we  may  compute  the 
population  at  any  future  time  on  the  supposition  of  a  uniform 
rate  of  increase. 

EXAMPLES. 

Ex.  1.  The  number  of  the  inhabitants  of  the  United  States 
in  1790  was  3,900,000,  and  in  1840,  17,000,000.  What  was 
the  average  increase  for  every  ten  years  ? 

Ans.  34  per  cent. 

Ex.  2.  Suppose  the  rate  of  increase  to  remain  the  same  for 
the  next  ten  years,  what  would  be  the  number  of  inhabitants 
in  1850? 

Ans.  22,800,000. 

Ex.  3.  At  the  same  rate,  in  what  time  would  the  number  in 
1840  be  doubled? 

Ans.  23.54  years. 
Ex.  4.  At  the  same  rate,  what  was  the  population  in  1780  ? 

Ans.  2,900,000. 

Ex.  5.  At  the  same  rate,  in  what  time  would  the  number  in 
1840  be  tripled  ? 

Ans.  37.31  years. 


MISCELLANEOUS  EXAMPLES. 


(356.)  Ex.  1.  Given  ^+-=0, 

11         I   to  find  the  values  of  z,  y, 
x    z       '  I        and  z. 
1     1 


2  2  2 

~a+&— c'  y~a—b+c'    ~b+c—a 

Ex.  2.  Given  x+y+z  +t  +u  =25, 


x+y+z  +t  -\-w=21,  I   to  find  the  values  of  x, 


x+y+u+t +w=28, 


y,  2,  t,  u,  and  w. 


#+z+w+£+w=29, 
y+z+u+t+w=30,) 

Ans.  x=3,  y=4,  2=5,  w=6,  £=7,  i«=8. 

fe.  3.  Given  x(x+y+z) =27,  \.     «    ,    , 

,1-      J  =  18.     to  find  the  values  of  ^.y, 


J.TIS.  x=3,  y=2,  2=4. 
Ba;.  4.  Given  xy=z,  ^ 

yz—v,   .  to  gn(j  tjie  vaiues  Of  ^  ^  2>  an(j  v< 

a;t)=a, 
y»=6a?J 

\/a 

Ex.  5.  Given  xyz  =  105,>j 

xyv=I35,  I   to  find  the  values  of  #,  y,  2,  and 

xzv=189,  f       v. 

y2v=315,J 

jins.  #=3,  y=5,  2=7,  v—9. 


326  MISCELLANEOUS    EXAMPLES. 

X* 

Ex.  6.  Given  x*+-+y'=84, 


to  find  the  values  of  x  and    . 


.  #=4,  y=2  or\8. 

7.  Given  -  --  =b}  to  find  the  values  of  x. 
a+</<z2-:ea 


no   x- 

14-6' 


i/x+Vx—a      ab* 

Ex.  8.  Given  -         =  -  ,  to  find  the  values  of  x. 
^/x-Vx—a    x~a 

a(l±b)> 
Ans.  x— 


Ex.  9.  Given  v/y—  «Ja—x—  Vy—x,  )  to  find  the  values 

2  Vy-a;+2  Va-a;=5  Va-x,  )     of  «  and  y- 


4          5 
-a,  y—~: 

D  Tt 


.  10.  Given  --  1  --  -  —  =  \/r,  to  find  the  values  of  x. 


=  \/r, 
v  b 


Ans.  x= 

Ex.  11.  Given  re8  +xy*=ay, 
y 


=ay,  )  e 

.     ,      [  to  find  the  values  of  ^  and  y. 

=bx,  ) 


,  . 


_  ., 

JEa:.  12.  Given  -  -  --  ---  =-  -  77-7—,  to  find 

a+x  a—x         3a—4b+x 

the  values  of  #. 

2a> 
Ans.  x=—3a,  or  3a  —  r-. 

a—x    b—x    a+b  f 

Ex.  13.  Given  -r-,  ---  ;  —  =  -  r>  to  find  the  values  of  #. 
o+x    a+x    a—b 

,  .? 


Ex.  14.  Given 


MISCELLANEOUS    EXAMPLES. 
Vl  _  X 


1+Vl+x    1—Vl—x 


327 

to  find  the  values  ofx. 


Ex.  15.  Given 


a+x 


Ans.  #= 

c 
=6,  to  find  the  values  of  x. 


.    X  —  ~~ 


=   10, 


tO  find  th® 


_ 
'+  Vy> 


values  of 
=275,  )      x  and  y. 

Ans.  x=9,  y=4. 

——f     to  find  the  values 
x+y    x+y  ±     ofa;and 

=41,     J 


^^=——   ( 


Ex.  16.  Given 


Ex.  17.  Given 


^x.  18.  Given  (a;+y)8+x+y=30,  )  to  find  the  values  of  x 
x—y=   1,  )      and  y. 

Ans.  x=2,  y=l. 

Ex.  19.  Given  x*—  4x*+7x*—  6z=18,  to  find  the  values  of  a; 
by  a  quadratic  equation. 

Ans.  x=3,  or  —1. 

Ex.  20.  Given  (x  +y  )  (x  y  +1)=   18x  y,  )  to  find  the  values 

+l)=2Q8 
Ans.  #= 


(x  +y  )  (x  y  +1)=   18x  y,  ) 
(x*+y*)  (xy+l)=2Q8xy,  \ 


of  x  and  y. 


E&.  21.  Given  (xa+y2)x?/=  13090,  )  to  find  the  values  of  x 
x+y       =18,         j      and  y. 

Ans.  x=7,    or  11, 

y=ll,  or  7. 

JBa;.  22.  Given  5(z2+?/2)+4:ry=356,  )  to  find  the  values  of 
x*+y*+x+y=62,    }      x  and  y. 

Ans.  x=4,  y=Q. 

Ex.  23.  Given  (x'+y*)xy=300,  )  to  find  the  values  of  a;  and 
x'+y*       =337,  \      y. 

Ans.  #=4,  y~3. 
15 


328  MISCELLANEOUS    EXAMPLES. 

Ex.  24.  Given  (x*+y*)  (aj'+y')=455,  )  to  find  the  values  of 
x+y  =5,      j      x  and  y. 

Ans.  x=3,  y=2. 

Ex.  25.  Given ,      y  +14,  ] 

x+y  I  to  find  the  values  of  a;  and 

and  — 


x-y 

#=12,  y=Q. 

to  find  the 


Ex.  26.  Given  (:ra— xy+y*)  («a+ya)         =91,    . 

(^-ay+y-)  " fc     ™  5     values  of 


Ans.  x—2,  or  —3;  y=3,  or  —2. 

fin 
and 


27.  Given  (x+y)xy    =30,    )  to  find  the  values  of  x 
:y  =468,  J 


Sr.  28.  The  sum  of  two  numbers  is  a,  and  the  sum  of  their 
reciprocals  is  b.     Required  the  numbers. 


a        /a*    a 

Ans-  *v  4- 


jE;r.  29.  In  the  composition  of  a  certain  quantity  of  gunpow- 
der, the  nitre  was  ten  pounds  more  than  two  thirds  of  the 
whole  ;  the  sulphur  was  four  and  a  half  pounds  less  than  one 
sixth  of  the  whole  ;  and  the  charcoal  was  two  pounds  less  than 
one  seventh  of  the  nitre.  How  many  pounds  of  gunpowder 
were  there  ? 

Ans.  69  pounds. 

Ex.  30.  Find  three  numbers  such  that  if  six  be  subtracted 
from  the  first  and  second,  the  remainders  will  be  in  the  ratio 
of  2  :  8  ;  if  thirty  be  added  to  the  first  and  third,  the  sums  will 
be  in  the  ratio  of  3  :  4  ;  but  if  ten  be  subtracted  from  the  sec- 
ond and  third,  the  remainders  will  be  as  4  :  5. 

Ans.  30,  42,  50. 

Ex.  31.  Divide  the  number  165  into  five  such  parts  that  the 
first  increased  by  one,  the  second  increased  by  two,  the  third 
diminished  by  three,  the  fourth  multiplied  by  4,  and  the  fifth 
divided  by  five,  may  all  be  equal. 

Ans    19,  18,  23,  5,  and  100. 


MISCELLANEOUS    EXAMPLES. 

Ex.  32.  A  criminal  having  escaped  from  prison,  traveled 
ten  hours  before  his  escape  was  known.  He  was  then  pur- 
sued, so  as  to  be  gained  upon  three  miles  an  hour.  After  his 
pursuers  had  traveled  eight  hours,  they  met  an  express  going 
at  the  same  rate  as  themselves,  who  met  the  criminal  two 
hours  and  twenty-four  minutes  before.  In  what  time  from  the 
commencement  of  the  pursuit  will  they  overtake  him  ? 

Ans.  20  hours. 

Ex.  33.  A  and  B  engage  to  reap  a  field  of  wheat  in  twelve 
days.  The  times  in  which  they  could  severally  reap  an  acre 
are  as  2  :  3.  After  some  days,  finding  themselves  unable  to 
finish  it  in  the  stipulated  time,  they  call  in  C  to  help  them, 
whose  rate  of  working  was  such  that,  if  he  had  wrought  with 
them  from  the  beginning,  it  would  have  been  finished  in  nine 
days.  Also,  the  times  in  which  he  could  have  reaped  the  field 
with  A  alone,  and  with  B  alone,  are  in  the  ratio  of  7  :  8. 
When  was  C  called  in  ? 

Ans.  After  six  days. 

Ex.  34.  A  laborer  is  engaged  for  n  days,  on  condition  that 
he  receives  p  pence  for  every  day  he  works,  and  pays  q  pence 
for  every  day  he  is  idle.  At  the  end  of  the  time  he  receives 
a  pence.  How  many  days  did  he  work,  and  how  many  was  he 
idle? 


Ans.  He  worked  -^-  —  ,  and  was  idle  -~  —  days. 
p+q  p+q 

Ex.  35.  The  fore  wheel  of  a  carriage  makes  three  revolu- 
tions more  than  the  hind  wheel  in  going  sixty  yards  ;  but,  if 
the  circumference  of  each  wheel  be  increased  one  yard,  it  will 
make  only  two  revolutions  more  than  the  hind  wheel  in  the 
same  space.  Required  the  circumference  of  each. 

Ans.  4  and  5  yards. 

Ex.  36.  There  is  a  wagon  with  a  mechanical  contrivance 
by  which  the  difference  of  the  number  of  revolutions  of  the 
wheels  on  a  journey  is  noted.  The  circumference  of  the  fore 
wheel  is  a  feet,  and  of  the  hind  wheel  b  feet.  What  is  the  dis- 
tance gone  over  when  the  fore  wheel  has  made  n  revolutions 
more  than  the  hind  wheel  ? 

abn 

Ans.  r  -  feet. 
6—  a 


330  MISCELLANEOUS    EXAMPLE*. 

Ex.  37.  A  merchant  has  two  casks,  each  containing  a  cer- 
tain quantity  of  wine.  In  order  to  have  an  equal  quantity  in 
each,  he  pours  out  of  the  first  cask  into  the  second  as  much 
as  the  second  contained  at  first ;  then  he  pours  from  the  second 
into  the  first  as  much  as  was  left  in  the  first ;  and  then  again. 
from  the  first  into  the  second  as  much  as  was  left  in  the  sec- 
ond, when  there  are  found  to  be  a  gallons  in  each  cask.  How 
many  gallons  did  each  cask  contain  at  first  ? 

lla       .  5« 
Ans.  —  and  — . 

Ex.  38.  A  and  B  engage  to  reap  a  field  for  $24 ;  and  as  A 
alone  could  reap  it  in  nine  days,  they  promise  to  complete  it  in 
five  days.  They  found,  however,  that  they  were  obliged  to 
call  in  C  to  assist  them  for  the  last  two  days,  in  consequence 
of  which  B  received  one  dollar  less  than  he  otherwise  would 
have  done.  In  what  time  could  B  or  C  alone  reap  the  field  ? 

Ans.  B  in  15,  and  C  in  18  days. 

Ex.  39.  A  cistern  can  be  filled  by  four  pipes  ;  by  the  first  in 
a  hours,  by  the  second  in  b  hours,  by  the  third  in  c  hours,  and 
by  the  fourth  in  d  hours.  In  what  time  will  the  cistern  be 
filled  when  the  four  pipes  are  opened  at  once  ? 

abed 

/\  77  *?       n~~i    ^ r  - 

abc+abd+acd+bcd' 

Ex.  40.  The  sum  of  the  cubes  of  two  numbers  is  35,  and 
the  sum  of  their  ninth  powers  is  20195.  Required  the  num- 
bers. 

Ans.  2  and  3. 

Ex.  41.  A  number  consisting  of  three  digits,  which  are  in 
Arithmetical  Progression,  being  divided  by  the  sum  of  its  dig- 
its, gives  a  quotient  26;  and  if  198  be  added  to  it,  the  digits 
will  be  inverted.  Required  the  number. 

Ans.  234. 

Ex.  42.  There  are  three  numbers  in  Geometrical  Progres- 
sion, the  difference  of  whose  differences  is  six,  and  their  sum  is 
forty-two.  Required  the  numbers. 

Ans.  6,  12,  and  24. 

Ex.  43.  There  are  three  numbers  in  harmonical  proportion ; 


MISCELLANEOUS    EXAMPLES.  331 

the  sum  of  the  first  and  third  is  18,  and  the  product  of  the  three 
numbers  is  576.     Required  the  numbers. 

Ans.  12,  8,  and  6. 

Ex.  44.  There  are  three  numbers  in  harmonica!  proportion, 
the  difference  of  whose  differences  is  2,  and  four  times  the 
product  of  the  first  and  third  is  960.  Required  the  numbers. 

Ans.  20,  15,  12. 

Ex.  45.  There  are  two  numbers  whose  product  is  300 ;  and 
the  difference  of  their  cubes  is  thirty-seven  times  the  cube  of 
their  difference.  What  are  the  numbers  ? 

Ans.  20  and  15. 

Ex.  46.  There  are  three  numbers  in  geometrical  progres- 
sion, the  greatest  of  which  exceeds  the  least  by  24 ;  and  the 
difference  of  the  squares  of  the  greatest  and  the  least  is  to  the 
sum  of  the  squares  of  all  the  three  numbers  as  5  :  7.  What 
are  the  numbers  ? 

Ans.  8,  16,  and  32. 

Ex.  47.  A  merchant  had  826,000,  which  he  divided  into  two 
parts,  and  placed  them  at  interest  in  such  a  manner  that  the 
incomes  from  them  were  equal.  If  he  had  put  out  the  first 
portion  at  the  same  rate  as  the  second,  he  would  have  drawn 
for  this  part  $720  interest ;  and  if  he  had  placed  the  second 
out  at  the  same  rate  as  the  first,  he  would  have  drawn  for  it 
$980  interest.  What  were  the  two  rates  of  interest  ? 

Ans.  6  per  cent,  for  the  larger  sum,  and  7  for  the  smaller. 

Ex.  48.  A  grocer  has  a  cask  containing  20  gallons  of  bran- 
dy, from  which  he  draws  off  a  certain  quantity  into  another 
cask  of  equal  size,  and,  having  filled  the  last  with  water,  the 
first  cask  was  filled  with  the  mixture.  It  now  appears  that 
if  6|  gallons  of  the  mixture  are  drawn  off  from  the  first  into 
the  second  cask,  there  will  be  equal  quantities  of  brandy  in 
each.  Required  the  quantity  of  brandy  first  drawn  off. 

Ans.  10  gallons. 

Ex.  49.  A  miner  bought  two  cubical  masses  of  ore  for  $820. 
Each  of  them  cost  as  many  dollars  per  cubic  foot  as  there  were 
feet  in  a  side  of  the  other ;  and  the  base  of  the  greater  con- 
tained a  square  yard  more  than  the  base  of  the  less.  What 
was  the  price  of  each  ? 

Ans.  500  and  320  dollars. 
15* 


MISCELLANEOUS    EXAMPLES. 

Ex.  50.  A  and  B  traveled  on  the  same  road,  and  at  the 
same  rate,  from  Cumberland  to  Baltimore.  At  the  50th  mile- 
stone from  Baltimore,  A  overtook  a  drove  of  geese,  which  were 
proceeding  at  the  rate  of  three  miles  in  two  hours ;  and  two 
hours  afterward  met  a  wagon,  which  was  moving  at  the  rate 
of  nine  miles  in  four  hours.  B  overtook  the  same  drove  of 
geese  at  the  45th  milestone,  and  met  the  same  wagon  40  min- 
utes before  he  came  to  the  31st  milestone.  Where  was  B 
when  A  reached  Baltimore  ? 

Ans.  25  miles  from  Baltimore. 

Ex.  51.  A  gentleman  bought  a  rectangular  lot  of  land  at  the 
rate  of  ten  dollars  for  every  foot  in  the  perimeter.  If  the 
same  quantity  had  been  in  a  square  form,  and  he  had  bought 
it  at  the  same  rate,  it  would  have  cost  him  $330  less ;  but  if 
he  had  bought  a  square  piece  of  the  same  perimeter,  he  would 
have  had  12  J  rods  more.  What  were  the  dimensions  of  the 
lot? 

Ans.  9  by  16  rods. 

Ex.  52.  A  and  B  put  out  at  interest  sums  amounting  to 
$2400.  A's  rate  of  interest  was  one  per  cent,  more  than  B's ; 
his  yearly  interest  was  five  sixths  of  B's ;  and  at  the  end  of 
ten  years  his  principal  and  simple  interest  amounted  to  five 
sevenths  of  B's.  What  sum  was  put  at  interest  by  each,  and 
at  what  rate  ? 

Ans.  A  $960,  at  5  per  cent. 
B  $1440,  at  4       " 

Ex.  53.  Two  merchants  sold  the  same  kind  of  cloth.  The 
second  sold  three  yards  more  of  it  than  the  first,  and  together 
they  received  $35.  The  first  said  to  the  second,  I  should  have 
received  $24  for  your  cloth  ;  the  other  replied,  I  should  have 
received  $12j  for  yours.  How  many  yards  did  each  of  them 
sell? 

Ans.  The  first  merchant,  5  or  15  yards. 
The  second     "        8  or  18      " 

Ex.  54.  A  person  bought  a  quantity  of  cloth  of  two  sorts  for 
$63.  For  every  yard  of  the  best  piece  he  gave  as  many  dol- 
lars as  he  had  yards  in  all ;  and  for  every  yard  of  the  poorer, 
as  many  dollars  as  there  were  yards  of  the  better  piece  more 
than  of  the  poorer.  Also,  the  whole  cost  of  the  best  piece  was 


MISCELLANEOUS    EXAMPLES.  333 

six  times  that  of  the  poorer.     How  many  yards  had  he  of 
each? 

Ans.  6  yards  of  the  better,  and  3  of  the  poorer. 

Ex.  55.  A  and  B,  165  miles  distant  from  each  other,  set  out 
with  a  design  to  meet.  A  travels  1  mile  the  first  day,  2  the 
second,  3  the  third,  and  so  on.  B  travels  20  miles  the  first 
day,  18  the  second,  16  the  third,  and  so  on.  In  how  many  days 
will  they  meet  ? 

Ans.  10  or  33  days. 

Ex.  56.  There  are  three  numbers  in  Geometrical  Progres- 
sion whose  continued  product  is  216,  and  the  sum  of  their 
cubes  is  1971.  Required  the  numbers. 

Ans.  3,  6,  and  12. 

Ex.  57.  There  are  four  numbers  in  Geometrical  Progression 
whose  sum  is  350 ;  and  the  difference  between  the  extremes  is 
to  the  difference  of  the  means  as  37  :  12.  What  are  the  num- 
bers? 

Ans.  54,  72,  96,  128. 

Ex.  58.  A  commences  a  piece  of  work  alone,  and  labors  for 
two  thirds  of  the  time  that  B  would  have  required  to  perform 
the  entire  work.  B  then  completes  the  job.  Had  both  labor- 
ed together,  it  would  have  been  completed  two  days  sooner ; 
and  A  would  have  performed  only  half  what  he  left  for  B. 
Required  the  time  in  which  they  would  have  performed  the 
work  separately. 

Ans.  A  in  6  days,  and  B  in  3  days. 

Ex.  59.  A  ship,  with  a  crew  of  175  men,  set  sail  with  a  sup- 
ply of  water  sufficient  to  last  to  the  end  of  the  voyage ;  but  in 
30  days  the  scurvy  made  its  appearance,  and  carried  off  three 
men  every  day ;  and  at  the  same  time  a  storm  arose  which 
protracted  the  voyage  three  weeks.  They  were,  however, 
just  enabled  to  arrive  in  port  without  any  diminution  in  each 
man's  daily  allowance  of  water.  Required  the  time  of  the 
passage,  and  the  number  of  men  alive  when  the  vessel  reached 
the  harbor. 

Ans.  The  voyage  lasted  79  days,  and  the  number  of  men 
alive  was  28. 

Ex.   60.    The    number   of  deaths   in   a   besieged   garrison 


334  MISCELLANEOUS    EXAMPLES. 

amounted  to  6  daily ;  and,  allowing  for  this  diminution,  their 
stock  of  provisions  was  sufficient  to  last  8  days.  But  on  the 
evening  of  the  sixth  day  100  men  were  killed  in  a  sally,  and 
afterward  the  mortality  increased  to  10  daily.  Supposing  the 
stock  of  provisions  unconsumed  at  the  end  of  the  sixth  day  to 
support  6  men  for  61  days,  it  is  required  to  find  how  long  it 
would  support  the  garrison,  arid  the  number  of  men  alive  when 
the  provisions  were  exhausted. 

Ans.  The  provisions  last  6  days,  and  26  men  survive. 


THE    END. 


C0omi01  bourse  ot  4iilcttl)ematu$, 

•  PUBLISHED    BT 

HARPER  &  BROTHERS,  NEW  YORK, 

THE  Publishers  of  the  course  of  Mathematics  by  Prof.  Loomis  invite  the  at- 
tention of  professors  of  colleges  and  teachers  generally  to  an  examination  ot 
these  works.  They  are  the  fruits  of  a  long  series  of  years  devoted  to  collegiate 
instruction,  and  it  is  believed  that  they  possess  in  an  eminent  degree  the  quali- 
ties of  simplicity,  conciseness,  and  lucid  arrangement,  and  are  adapted  to  the 
wants  of  students  generally  in  our  colleges  and  academies. 

LOOMIS'  TREATISE  ON  ALGEBRA. 

8vo,  p.  334,  Sheep,  $1  00.     Third  Edition. 

I  hare  carefully  examined  the  work  of  Prof.  Loomis  on  Algebra,  and  am  much  pleased  with  it. 
The  arrangement  is  sufficiently  scientific,  yet  the  order  of  the  topics  is  obviously,  and,  I  think,  ju- 
diciously made  with  reference  to  the  development  of  the  powers  of  the  pupil.  The  most  rigorous 
modes  of  reasoning  are  designedly  avoided  in  the  earlier  portions  of  the  work,  and  deferred  till  the 
student  is  better  fitted  to  appreciate  them.  All  the  principles  are,  however,  established  with  suffi- 
cient rigor  to  give  satisfaction.  Much  care  seems  to  have  been  taken,  by  generalizing  particular  ex- 
amples and  other  means,  to  develop  the  faculty  and  induce  the  habit  of  generalizing,  a  point  which- 
I  think,  has  not  received  sufficient  attention  hitherto.  On  the  whole,  therefore,  I  think  this  work 
better  suited  for  the  purposes  of  a  text-book  than  any  other  I  have  Been. — AUGUSTUS  W.  SMITH 
LL.D.,  Professor  of  Mathematics  and  Astronomy  in  the  Wesleyan  University. 

Prof.  Loomi*'  Treatise  on  Algebra  is  an  excellent  elementary  work.  It  is  sufficiently  extensive 
for  ordinary  purposes,  and  is  characterized  throughout  by  a  happy  combination  of  brevity  and  clear- 
ness.— A.  CASWELL,  D.D.,  Professor  of  Mathematics  and  Natural  Philosophy  in  Brown  University. 

I  have  examined  Prof.  Loomis'  new  work  on  Algebra,  and  am  highly  pleased  with  it.  For  con 
ciseness  and  clearness  of  statement,  and  for  its  lucid  explanation  of  elementary  principles,  it  is  de 
cidedly  superior  to  any  work  with  which  I  am  acquainted.  I  hope  it  will  be  extensively  used  in  all 
our  public  institutions. — ALONZO  GRAY,  A.M.,  Professor  in  Brooklyn  Female  Academy. 

I  have  examined  Prof.  Loomis'  Algebra  carefully  and  with  much  interest,  and  am  so  perfectly 
satisfied  with  it,  that  I  shall  introduce  it  to  my  classes  as  soon  as  possible.  It  is  just  the  work  which 
I  have  been  for  a  long  time  in  search  of.  I  am  particularly  delighted  with  the  mode  of  treating  the 
subject  of  logarithms,  and,  indeed,  with  the  clearness  of  the  investigations  generally  throughout  the 
work. — E.  OTIS  KENDALL,  Professor  of  Mathematics  and  Astronomy  in  the  Central  High  School  of 
Philadelphia. 

I  fully  concur  with  Prof.  Kendall  in  his  opinion  of  Loony's'  Algebra.— SEARS  C.  WALKER,  of  the 
United  States  Coast  Survey. 

A  text-book  like  this  of  Prof.  Loomis  was  much  needed,  and  the  desideratum  is  so  well  supplied, 
f  hat  I  think  it  can  not  fail  to  commend  itself  at  once  to  the  favorable  regard  of  those  who  are  looking 
for  the  best  work  for  college  classes.  I  consider  it  decidedly  the  best  book  for  college  instruction 
that  I  am  acquainted  with  on  the  subject,  and  it  has  been  adopted  as  a  text-book  in  our  college  by 
unanimous  consent  of  the  faculty.  Prof.  Loomis  has  been  very  happy  in  simplifying  the  more  diffi- 
cult parts  of  the  subject,  especially  on  the  theory  of  equations  and  on  logarithms.— JAMES  NOONEY, 
A.M.,  Professor  of  Mathematics  and  Natural  Philosophy  in  Western  Reserve  College. 

I  have  carefully  examined  Prof.  Loomis'  Algebra,  and  think  it  better  adapted  for  a  text-book  lor 
college  students  than  any  other  I  have  seen.— C.  GILL,  Professor  of  Mathematics  in  St.  Pants  Col. 

Prof.  Loomis  seems  very  happily  to  have  observed  the  proper  medium  between  exuberance  of  ex- 
planation and  demonstration  on  the  one  hand,  which  leaves  little  or  nothing  for  the  student  himself 
K,  do  ;  and  a  repulsive  conciseness  on  the  other,  which  discourages  him,  and  gives  him  a  disrelish 
fcr  this  portion  of  study.  I  have  adopted  it  as  a  text-book  in  the  Cornelius  Institute,  believing  it  to 
be  better  suited  to  youth  who  are  preparing  for  college,  than  any  other  treatise  on  Algebra  with 
waich  1  am  acquainted.— JOHN  J.  OWEN,  D.D.,  Professor  in  the  Free  Academy. 

After  a  thorough  examination  of  Prof.  Loomis'  work  on  Algebra,  I  have  concluded  to  adopt  it  as  a 
M  t-book  in  this  Institution.— MARCUS  CATLIN,  A.M.,  Professor  of  Mathematics  and  Astronomy  tw 
23  -milton  College 


2  Critical  Notices  of  Loomis'  Algebra. 

-    • .  -»  —  '•  .1  JE  >?»?  ^w**    •*   *  v- 

Prof.  Loomis'  work  on  Algebra  is  exceedingly  wall  adapted  for  the  purposes  of  instruction.  He 
has  avoided  the  difficulties  which  result  from  too  great  conciseness,  and  aiming  at  the  utmost  rigor 
of  demonstration ;  and,  at  the  same  time,  has  furnished  in  his  book  a  good  and  sufficient  preparation 
for  the  subsequent  parts  of  the  mathematical  course.  I  do  not  know  of  a  treatise  which,  all  things 
considered,  keeps  both  these  objects  so  steadily  in  view.— J.  WARD  ANDREWS,  A.M.,  Professor  of 
Mathematics  and  Natural  Philosophy  in  Marietta  College. 

Prof.  Loomis'  work  is  well  calculated  to  impart  a  clear  and  correct  knowledge  of  the  principles  of 
Algebra.  The  rules  are  concise,  yet  sufficiently  comprehensive,  containing  in  few  words  all  that  is 
necessary,  and  nothing  more;  the  absence  of  which  quality  mars  many  a  scientific  treatise.  The 
collection  of  problems  is  peculiarly  rich,  adapted  to  impress  the  most  important  principles  upon  the 
youthful  mind,  and  the  student  is  led  gradually  and  intelligently  into  the  more  interesting  and  higher 
departments  of  the  science.— JOHN  BROCKLESBY,  A.M.,  Professor  of  Mathematics  and  Natural 
Philosophy  in  Trinity  College. 

I  am  much  pleased  with  Prof.  Loomis'  Algebra.  I  think  he  has  accomplished  very  happily  the 
object  he  had  in  view,  and  has  prepared  a  work  remarkably  well  adapted  for  the  use  of  college  stu 
dents.— E.  S.  SHELL,  A.M.,  Professor  of  Mathematics  and  Natural  Philosophy  in  Amherst  College 

I  am  much  pleased  with  Prof.  Loomis'  Algebra.  The  arrangement  of  the  subjects  Is,  I  think,  an 
admirable  one.  The  best  proof  I  can  give  of  the  estimation  in  which  I  hold  it  is,  that  I  am  now 
hearing  recitations  in  it  the  second  time.— JOHN  TATLOCK,  A.M.,  Professor  of  Mathematics  in 
Williams  College. 

I  have  examined  Prof.  Loomis'  Algebra  with  great  attention,  and  am  so  well  pleased  with  its  ar- 
larigement  and  execution  throughout,  that  I  have  decided  to  adopt  it  as  a  text-book  in  this  institu* 
tion.— THOMAS  E.  SUDLKR,  A.M.,  Professor  of  Mathematics  in  Dickinson  College. 

Prof.  Loomis  has  here  aimed  at  exhibiting  the  first  principles  of  Algebra  in  a  form  which,  while 
evel  with  the  capacities  of  ordinary  students  and  the  present  state  of  the  science,  is  fitted  to  elicit 
that  degree  of  effort  which  educational  purposes  require.  Throughout  the  work,  whenever  it  can 
be  done  with  advantage,  the  practice  is  followed  of  generalizing  particular  examples,  or  of  extend- 
ing a  question  proposed  relative  to  a  particular  quantity,  to  the  class  of  quantities  to  which  it  be- 
longs ;  a  practice  of  obvious  utility,  as  accustoming  the  student  to  pass  from  the  particular  to  the 
general,  and  as  fitted  to  impress  a  main  distinction  between  the  literal  and  numeral  calculus.  The 
general  doctrine  of  Equations  is  expounded  with  clearness,  and,  we  may  add,  with  independence. 
The  author  has  developed  this  subject  in  an  order  of  his  own.  Theorems  which  find  a  place  in  other 
treatises  are  omitted,  and  what  sometimes  appears  in  a  generic  form,  or  in  that  of  a  corollary,  be- 
comes specific,  or  assumes  the  place  of  a  primary  proposition.  We  venture  to  say  that  there  will 
be  but  one  opinion  respecting  the  general  character  of  the  exposition. — American  Journal  of  Science 
and  Arts. 

I  regard  Prof.  Loomis'  Algebra  as  altogether  worthy  of  the  high  reputation  its  author  deservedly 
enjoys.  It  possesses  those  qualities  wl^ch  are  chiefly  requisite  in  a  college  text-book.  Its  state- 
ments are  clear  and  definite ;  the  more  important  principles  are  made  so  prominent  as  to  arrest  the 
pupil's  attention  ;  and  it  conducts  the  pupil  by  a  sure  and  easy  path  to  those  habits  of  generalization 
which  the  teacher  of  Algebra  has  so  much  difficulty  in  imparting  to  his  pupils.— JULIAN  M.  STUR- 
TEVANT,  L.L.D.,  President  of  Illinois  College. 

The  arrangement  of  Prof.  Loomis'  Algebra  is  good  i  the  doctrine  of  Equations  is  clearly  presented, 
and  the  principle  of  generalization  is  ably  developed  in  a  manner  well  calculated  to  improve  th» 
youthful  mind.— W.  P.  ALRICH,  Professor  of  Mathematics  in  Washington  College,  Pennsylvania. 

Prof.  Loomis'  Algebra  is  admirably  got  up.  It  is  clear  and  simple  in  arrangement,  and  just  the 
work  for  the  class  of  learners  for  whom  the  author  prepared  it.  The  introduction  of  Horner's  ad- 
mirable method  for  finding  incoiraiensurable  roots,  and  the  section  on  Logarithms,  render  it  superior 
to  any  text-book  on  the  subjec;  with  which  I  am  acquainted.— Pres.  COLLINS,  Emory  and  Henry 
College,  Virginia. 

We  feel  bound  to  express  our  conviction  that  this  is  a  decidedly  better  text-book,  especially  for 
those  not  already  far  advanced  in  the  study,  than  any  other  we  have  seen.  It  is  carefully  and  lu- 
cidly arranged,  and  admirably  enunciated  and  explained. — Teacher's  Journal. 

The  present  work  is  the  fruit  of  long  experience  in  teaching  and  diligent  investigation  of  the  sci 
race.  The  author  has  sought  to  avoid  unnecessary  prolixity  on  the  one  hand,  and  undue  brevity  o» 
the  other,  and  with  the  observance  of  this  happy  medium  he  has  embodied  all  the  latest  itnpror* 
mants.- Methodist  Quarterly  Rtview. 


Critical  Notices  of  Loomis'  Geometry.  3 

LOOMIS'  GEOMETRY  AND  CONIC  SECTIONS. 

Second  Edition.    8vo,  p.  226,  Sheep,  $1  00. 

Every  page  of  this  book  bears  marks  of  careful  preparation.  Only  those  propositions  are  selected 
which  are  most  important  in  themselves,  or  which  are  indispensable  in  the  demonstration  of  others. 
The  propositions  are  all  enunciated  with  studied  precision  and  brevity.  The  demonstrations  are 
complete  without  being  encumbered  with  verbiage  ;  and,  unlike  many  works  we  could  mention,  the 
diagrams  are  good  representations  of  the  objects  intended.  We  believe  this  book  will  take  its  place 
among  the  best  elementary  works  which  our  country  has  produced. — American  Review. 

Prof.  Loomis'  Geometry  is  characterized  by  the  same  neatness  and  elegance  which  were  exhibited 
in  his  Algebra.  While  tlie  logical  form  of  argumentation  peculiar  to  Playfair's  Euclid  is  preserved, 
more  completeness  and  symmetry  is  secured  by  additions  in  solid  and  spherical  geometry,  and  by  a 
different  arrangement  of  the  propositions.  It  will  be  a  favorite  with  those  who  admire  the  chaste 
forms  of  argumentation  of  the  old  school ;  and  it  is  a  question  whether  these  are  not  the  best  for  the 
purposes  of  mental  discipline. — Northern  Christian  Advocate. 

As  a  text-book  for  instruction,  this  work  possesses  advantages  superior,  in  some  respects,  to  any 
other  work  on  the  subject  in  our  language.  The  arrangement  is  good,  and  the  selection  of  proposi- 
tions so  judiciously  made  as  to  comprise  what  is  most  valuable  for  the  purposes  of  the  student,  both 
for  intellectual  culture  and  for  a  knowledge  of  geometry.  Prof.  Loomis  has  introduced  some  valu- 
able improvements,  especially  that  of  computing  the  area  of  a  circle  in  a  very  simple  and  easy  man 
ner,  and  that  of  shading  the  diagrams  in  solid  geometry,  which  will  greatly  aid  the  student  inform- 
ing his  conceptions  of  solid  angles  and  the  positions  of  planes. — Ohio  Observer. 

The  enunciations  in  Prof.  Loomis'  Geometry  are  concise  and  clear,  and  the  processes  neither  too 
brief  nor  too  diffuse.  The  part  treating  of  solid  Geometry  is  undoubtedly  superior,  in  clearness  and 
arrangement,  to  any  other  elementary  treatise  among  us. — New  York  Evangelist. 

Prof.  Loomis  has  admirably  harmonized  the  logical  system  of  the  Greek  geometer  with  the  more 
rapid  processes  of  modern  mathematicians. — New  York  Observer. 

Having  been  requested,  by  a  resolution  of  our  Board  of  Trustees,  to  report  such  a  course  of  mathe 
matical  studies  as  I  may  deem  best  suited  to  our  circumstances,  I  have  selected  Loomis'  Geometry 
and  Conic  Sections  as  a  part  of  the  course. — MATTHEW  J.  WILLIAMS,  Professor  of  Mathematics 
in  South  Carolina  College. 

Prof.  Loomis  has  made  Legendre's  Geometry  far  more  Euclidian,  and  therefore  more  valuable. 
Some  of  his  demonstrations  are  decided  improvements  on  those  of  Legendre.—  Professor  C.  DEWET, 
Rochester  Collegiate  Institute. 

This  book  is  far  in  advance  of  Playfair's  Euclid.    It  can  not  fail  to  come  into  general  use. — Albany 

Atlas. 

Particular  attention  has  been  paid  by  the  author  to  the  diagrams,  reducing  them  to  a  uniformity 
of  dimensions,  and,  of  course,  conveying  clearer  ideas  of  what  they  represent  to  the  mind  of  the 
pupil.  This  book  is  worthy  the  attention  of  our  high  schools  and  colleges. — Hartford  Spectator. 

The  propositions  are  all  enunciated  with  clearness  and  brevity.  It  is  a  valuable  work,  and  well 
deserving  the  attention  of  schools  and  mechanics. — Farmer  and  Mechanic. 

An  available  text-book,  which  we  have  no  doubt  will  become  a  standard  authority  in  this  depart- 
ment of  instruction. — Literary  World. 

This  work  is  admirably  drawn  up,  and  merits  universal  adoption  in  all  schools  where  these  branches 
of  science  are  taught.— Courier  and  Enquirer. 

These  books  are  terse  in  style,  clear  in  method,  easy  of  comprehension,  and  perfectly  free  from 
that  useless  verbiage  with  which  it  is  too  much  the  fashion  to  load  school-books  under  prctonse  of 
explanation. — Scott's  Weekly  Paper,  Canada. 

Prof.  Loomis  is  doing  a  valuable  service  to  the  cause  of  mathematical  science  by  the  course  of 
text-books  he  is  preparing.  His  writings  in  other  departments  of  science  are  characterized  by  a 
remarkable  clearness  in  the  manner  in  which  he  exhibits  truth,  and  his  treatises  on  Algebra  and 
Geometry  bear  evident  marks  of  having  emanated  from  the  same  mind. — JAMES  H.  COFFIN,  A.M., 
Professor  of  Mathematict  in  Lafayette  College 

Prof.  Loomis  has  made  many  improvements  in  Legendre's  Geometry,  retaining  all  the  merits  of 
that  author  without  the  defects.  I  shall  adopt  his  work  as  a  text-book  in  this  college.— THOMAS  E 
SUDI.ER,  A.M.,  Professor  of  Mathematict  in  Dickinson  College. 


4    Critical  Notices  of  Loomis*  Trigonometry  and  Tables. 
LOOMIS'  TRIGONOMETRY  AND  TABLES. 

8vo,  p.  320,  Sheep,  $1  50. 

This  work  contains  an  exposition  of  the  nature  and  properties  of  Logarithms , 
the  principles  of  Plane  Trigonometry ;  the  Mensuration  of  Surfaces  and  Solids ; 
the  principles  of  Land  Surveying,  with  a  full  description  of  the  instruments  em- 
ployed ;  the  Elements  of  Navigation,  and  of  Spherical  Trigonometry.  The 
Tables  furnish  the  Logarithms  of  Numbers  to  10,000  ;  logarithmic  Sines  and 
Tangents  for  every  ten  Seconds  of  the  Quadrant ;  natural  Sines  and  Tangents 
for  every  Minute  of  the  Quadrant ;  a  Traverse  Table ;  a  Table  of  Meridional 
Parts,  &c. 

The  following  are  a  few  of  the  notices  of  this  work  which  have  been  received 
by  the  publishers. 

Loomis'  Trigonometry  is  well  adapted  to  give  the  student  that  distinct  knowledge  of  the  princi- 
ples of  the  science  so  important  in  the  further  prosecution  of  the  study  of  mathematics.  The  de- 
scription and  representation  of  the  instruments  used  in  surveying,  leveling,  etc.,  are  sufficient  to 
prepare  the  student  to  make  a  practical  application  of  the  principles  he  has  learned.  The  Tables 
are  just  the  thing  for  college  students.— JOHN  TATLOCK,  A.M.,  Professor  of  Mathematics  in  Wil- 
liams College. 

Prof.  Loomis  has  done  up  the  work  admirably.  The  brevity  and  clearness  which  characterize 
this  excellent  system  of  mathematical  reasoning  are  the  ne  plus  ultra  for  such  a  work.  His  Trigo- 
nometry will  meet  with  the  approval  already  accorded  to  his  Algebra  and  Geometry. — Professor  C. 
DEWEY,  Rochester  Collegiate  Institute. 

Loomis'  Trigonometry  is  sufficiently  extensive  for  collegiate  purposes,  and  is  every  where  clear 
and  simple  in  its  statements,  without  being  redundant.  The  learner  will  here  find  what  he  really 
needs  without  being  distracted  by  what  is  superfluous  or  irrelevant. — A.  CASWELL,  D.D.,  Professor 
of  Mathematics  and  Natural  Philosophy  in  Brown  University. 

Loomis'  Tables  are  vastly  better  than  those  in  common  use.  The  extension  of  the  sines  and  tan 
gents  to  ten  seconds  is  a  great  improvement.  The  tables  of  natural  sines  are  indispensable  to  a  good 
understanding  of  Trigonometry  ;  and  the  natural  tangents  are  exceedingly  convenient  in  analytical 
geometry. — J.  WARD  ANDREWS,  A.M.,  Professor  of  Mathematics  and  Natural  Philosophy  in  Mari- 
etta College. 

Loomis'  Trigonometry  and  Tables  are  a  great  acquisition  to  mathematical  schools.  I  know  of  no 
work  in  which  the  principles  of  Trigonometry  are  so  well  condensed  and  so  admirably  adapted  to 
the  course  of  instruction  in  the  mathematical  schools  of  our  country.  I  shall  adopt  it  as  a  text- 
book for  instruction  in  this  college.— THOMAS  E.  SUDLER,  A.M.,  Professor  of  Mathematics  in  Dick- 
inson College. 

Loomis'  Trigonometry  and  Tables  are  both  excellent  works,  and  I  shall  recommend  them  at  every 
opportunity  which  offers. — JAMES  CURLEY,  Professor  in  Georgetown  College. 

I  am  so  much  pleased  with  Prof.  Loomis'  Trigonometry  that  I  intend  to  use  it  as  a  text-book  in 
this  college.— JOHN  BROCKLESBY,  A.M.,  Professor  of  Mathematics  in  Trinity  College. 

In  this  work  the  principles  of  Trigonometry  and  its  applications  are  discussed  with  the  same  clear 
ness  that  characterizes  the  previous  volumes.  The  portion  appropriated  to  Mensuration,  Surveying 
&c.,  will  especially  commend  itself  to  teachers,  by  the  judgment  exhibited  in  the  extent  to  whicl 
they  are  carried,  and"  the  practically  useful  character  of  the  matter  introduced.  What  I  have  par 
icularly  admired  in  this,  as  well  as  the  previous  volumes,  is  the  constant  recognition  of  the  difficul 
ties,  present  and  prospective,  which  are  likely  to  embarrass  the  learner,  and  the  skill  and  tact  with 
which  they  are  removed.  The  Logarithmic  Tables  will  be  found  unsurpassed  in  practical  conven- 
ience by  any  others  of  the  same  extent. — AUGUSTUS  W.  SMITH,  LL.D.,  Professor  of  Mathematics 
and  Astronomy  in  the  Wesleyan  University. 

Prof.  Loomis'  text-books  in  Mathematics  are  models  of  neatness,  precision,  and  practical  adaptation 
to  the  wants  of  students. — Methodist  Quarterly  Review. 


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